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Problem 2. Let $a, b$ and $c$ be positive real numbers. Prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+\frac{8}{(c+a)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$
and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$.
Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b c} \geq \frac{4}{\left(a^{2}+b^{2}\right)(c+1)}
$$
Using the AM-GM inequality, we have
$$
\frac{4}{\left(a^{2}+b^{2}\right)(c+1)}+\frac{a^{2}+b^{2}}{2} \geq 2 \sqrt{\frac{2}{c+1}}=\frac{4}{\sqrt{2(c+1)}}
$$
respectively
$$
\frac{c+3}{8}=\frac{(c+1)+2}{8} \geq \frac{\sqrt{2(c+1)}}{4}
$$
We conclude that
$$
\frac{4}{\left(a^{2}+b^{2}\right)(c+1)}+\frac{a^{2}+b^{2}}{2} \geq \frac{8}{c+3}
$$
and finally
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(a+c)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem 1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0
$$
implying
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}
$$
It follows that $(a+b)(b+c)(c+a)=0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $0=2+3$, a contradiction). Thus $a+b+c+d \neq 0$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem 3. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold?
|
Solution 1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right) \\
& =\left(a b+1+\frac{a}{c}+a\right)+\left(b c+1+\frac{b}{a}+b\right)+\left(c a+1+\frac{c}{b}+c\right) \\
& =a b+b c+c a+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+3+a+b+c
\end{aligned}
$$
Notice that by AM-GM we have $a b+\frac{b}{a} \geq 2 b, b c+\frac{c}{b} \geq 2 c$, and $c a+\frac{a}{c} \geq 2 a$.
Thus ,
$\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq\left(a b+\frac{b}{a}\right)+\left(b c+\frac{c}{b}\right)+\left(c a+\frac{a}{c}\right)+3+a+b+c \geq 3(a+b+c+1)$.
The equality holds if and only if $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## Problem 2.
Prove that for all non-negative real numbers $x, y, z$, not all equal to 0 , the following inequality holds
$$
\frac{2 x^{2}-x+y+z}{x+y^{2}+z^{2}}+\frac{2 y^{2}+x-y+z}{x^{2}+y+z^{2}}+\frac{2 z^{2}+x+y-z}{x^{2}+y^{2}+z} \geqslant 3
$$
Determine all the triples $(x, y, z)$ for which the equality holds.
|
Solution. Let us first write the expression $L$ on the left hand side in the following way
$$
\begin{aligned}
L & =\left(\frac{2 x^{2}-x+y+z}{x+y^{2}+z^{2}}+2\right)+\left(\frac{2 y^{2}+x-y+z}{x^{2}+y+z^{2}}+2\right)+\left(\frac{2 z^{2}+x+y-z}{x^{2}+y^{2}+z}+2\right)-6 \\
& =\left(2 x^{2}+2 y^{2}+2 z^{2}+x+y+z\right)\left(\frac{1}{x+y^{2}+z^{2}}+\frac{1}{x^{2}+y+z^{2}}+\frac{1}{x^{2}+y^{2}+z}\right)-6
\end{aligned}
$$
If we introduce the notation $A=x+y^{2}+z^{2}, B=x^{2}+y+z^{2}, C=x^{2}+y^{2}+z$, then the previous relation becomes
$$
L=(A+B+C)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)-6
$$
Using the arithmetic-harmonic mean inequality or Cauchy-Schwartz inequality for positive real numbers $A, B, C$, we easily obtain
$$
(A+B+C)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right) \geqslant 9
$$
so it holds $L \geqslant 3$.
The equality occurs if and only if $A=B=C$, which is equivalent to the system of equations
$$
x^{2}-y^{2}=x-y, \quad y^{2}-z^{2}=y-z, \quad x^{2}-z^{2}=x-z
$$
It follows easily that the only solutions of this system are
$(x, y, z) \in\{(t, t, t) \mid t>0\} \cup\{(t, t, 1-t) \mid t \in[0,1]\} \cup\{(t, 1-t, t) \mid t \in[0,1]\} \cup\{(1-t, t, t) \mid t \in[0,1]\}$.
PSC Remark We feel the equality case needs more explanations in order to have a complete solution, our suggestion follows:
Clearly if $x, y, z$ are all equal and not 0 satisfy the condition. Now suppose that not all of them are equal it means we can't simultaneously have $x+y=y+z=z+x=1$ otherwise we would have all $x, y, z$ equal to $\frac{1}{2}$ which we already discussed. We can suppose now that $x=y$ and $y+z=z+x=1$ where we get $z=1-x$. So, all triples which satisfy the equality are $(x, y, z)=(a, a, a),(b, b, 1-b)$ and all permutations for any $a>0$ and $b \in[0,1]$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem 2. Let $x, y, z$ be positive integers such that $x \neq y \neq z \neq x$. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold?
|
Solution. Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
which are equivalent to
$$
x^{2}+y^{2} \geq 2 x y+1, \quad y^{2}+z^{2} \geq 2 y z+4, \quad x^{2}+z^{2} \geq 2 x z+9
$$
or otherwise
$$
z x^{2}+z y^{2} \geq 2 x y z+z, \quad x y^{2}+x z^{2} \geq 2 x y z+4 x, \quad y x^{2}+y z^{2} \geq 2 x y z+9 y
$$
Adding up the last three inequalities we have
$$
x y(x+y)+y z(y+z)+z x(z+x) \geq 6 x y z+4 x+9 y+z
$$
which implies that $(x+y+z)(x y+y z+z x-2) \geq 9 x y z+2 x+7 y-z$.
Since $x \geq z+3$ it follows that $2 x+7 y-z \geq 0$ and our inequality follows.
Case 2. $y=z+1$. Since $x \geq y+1=z+2$ it follows that $x \geq z+2$, and replacing $y=z+1$ in the required inequality we have to prove
$$
(x+z+1+z)(x(z+1)+(z+1) z+z x-2) \geq 9 x(z+1) z
$$
which is equivalent to
$$
(x+2 z+1)\left(z^{2}+2 z x+z+x-2\right)-9 x(z+1) z \geq 0
$$
Doing easy algebraic manipulations, this is equivalent to prove
$$
(x-z-2)(x-z+1)(2 z+1) \geq 0
$$
which is satisfied since $x \geq z+2$.
The equality is achieved only in the Case 2 for $x=z+2$, so we have equality when $(x, y, z)=$ $(k+2, k+1, k)$ and all the permutations for any positive integer $k$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A1. Let $a, b, c$ be positive real numbers such that $a+b+c+a b+b c+c a+a b c=7$. Prove that
$$
\sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \geq 6
$$
|
Solution. First we see that $x^{2}+y^{2}+1 \geq x y+x+y$. Indeed, this is equivalent to
$$
(x-y)^{2}+(x-1)^{2}+(y-1)^{2} \geq 0
$$
Therefore
$$
\begin{aligned}
& \sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \\
\geq & \sqrt{a b+a+b+1}+\sqrt{b c+b+c+1}+\sqrt{c a+c+a+1} \\
= & \sqrt{(a+1)(b+1)}+\sqrt{(b+1)(a+1)}+\sqrt{(c+1)(a+1)}
\end{aligned}
$$
It follows from the AM-GM inequality that
$$
\begin{aligned}
& \sqrt{(a+1)(b+1)}+\sqrt{(b+1)(a+1)}+\sqrt{(c+1)(a+1)} \\
\geq & 3 \sqrt[3]{\sqrt{(a+1)(b+1)} \cdot \sqrt{(b+1)(a+1)} \cdot \sqrt{(c+1)(a+1)}} \\
= & 3 \sqrt[3]{(a+1)(b+1)(c+1)}
\end{aligned}
$$
On the other hand, the given condition is equivalent to $(a+1)(b+1)(c+1)=8$ and we get the desired inequality.
Obviously, equality is attained if and only if $a=b=c=1$.
Remark. The condition of positivity of $a, b, c$ is superfluous and the equality $\cdots=7$ can be replaced by the inequality $\cdots \geq 7$. Indeed, the above proof and the triangle inequality imply that
$$
\begin{aligned}
\sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} & \geq 3 \sqrt[3]{(|a|+1)(|b|+1)(|c|+1)} \\
& \geq 3 \sqrt[3]{|a+1| \cdot|b+1| \cdot|c+1|} \geq 6
\end{aligned}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A3. Let $a, b, c, d$ be real numbers such that $0 \leq a \leq b \leq c \leq d$. Prove the inequality
$$
a b^{3}+b c^{3}+c d^{3}+d a^{3} \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}
$$
|
Solution. The inequality is equivalent to
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)^{2} \geq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\right)^{2}
$$
By the Cauchy-Schwarz inequality,
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right) \geq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\right)^{2}
$$
Hence it is sufficient to prove that
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)^{2} \geq\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right)
$$
i.e. to prove $a b^{3}+b c^{3}+c d^{3}+d a^{3} \geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$.
This inequality can be written successively
$$
a\left(b^{3}-d^{3}\right)+b\left(c^{3}-a^{3}\right)+c\left(d^{3}-b^{3}\right)+d\left(a^{3}-c^{3}\right) \geq 0
$$
or
$$
(a-c)\left(b^{3}-d^{3}\right)-(b-d)\left(a^{3}-c^{3}\right) \geq 0
$$
which comes down to
$$
(a-c)(b-d)\left(b^{2}+b d+d^{2}-a^{2}-a c-c^{2}\right) \geq 0
$$
The last inequality is true because $a-c \leq 0, b-d \leq 0$, and $\left(b^{2}-a^{2}\right)+(b d-a c)+\left(d^{2}-c^{2}\right) \geq 0$ as a sum of three non-negative numbers.
The last inequality is satisfied with equality whence $a=b$ and $c=d$. Combining this with the equality cases in the Cauchy-Schwarz inequality we obtain the equality cases for the initial inequality: $a=b=c=d$.
Remark. Instead of using the Cauchy-Schwarz inequality, once the inequality $a b^{3}+b c^{3}+c d^{3}+$ $d a^{3} \geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$ is established, we have $2\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right) \geq\left(a b^{3}+b c^{3}+c d^{3}+\right.$ $\left.d a^{3}\right)+\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right)=\left(a b^{3}+a^{3} b\right)+\left(b c^{3}+b^{3} c\right)+\left(c d^{3}+c^{3} d\right)+\left(d a^{3}+d^{3} a\right) \stackrel{A M-G M}{\geq}$ $2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} d^{2}+2 d^{2} a^{2}$ which gives the conclusion.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A4. Let $x, y, z$ be three distinct positive integers. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold?
|
Solution. Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
which are equivalent to
$$
x^{2}+y^{2} \geq 2 x y+1, \quad y^{2}+z^{2} \geq 2 y z+4, \quad x^{2}+z^{2} \geq 2 x z+9
$$
or otherwise
$$
z x^{2}+z y^{2} \geq 2 x y z+z, \quad x y^{2}+x z^{2} \geq 2 x y z+4 x, \quad y x^{2}+y z^{2} \geq 2 x y z+9 y
$$
Adding up the last three inequalities we have
$$
x y(x+y)+y z(y+z)+z x(z+x) \geq 6 x y z+4 x+9 y+z
$$
which implies that $(x+y+z)(x y+y z+z x-2) \geq 9 x y z+2 x+7 y-z$.
Since $x \geq z+3$ it follows that $2 x+7 y-z \geq 0$ and our inequality follows.
Case 2. $y=z+1$. Since $x \geq y+1=z+2$ it follows that $x \geq z+2$, and replacing $y=z+1$ in the required inequality we have to prove
$$
(x+z+1+z)(x(z+1)+(z+1) z+z x-2) \geq 9 x(z+1) z
$$
which is equivalent to
$$
(x+2 z+1)\left(z^{2}+2 z x+z+x-2\right)-9 x(z+1) z \geq 0
$$
Doing easy algebraic manipulations, this is equivalent to prove
$$
(x-z-2)(x-z+1)(2 z+1) \geq 0
$$
which is satisfied since $x \geq z+2$.
The equality is achieved only in the Case 2 for $x=z+2$, so we have equality when $(x, y, z)=$ $(k+2, k+1, k)$ and all the permutations for any positive integer $k$.
## Combinatorics
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem G5. A point $P$ lies in the interior of the triangle $A B C$. The lines $A P, B P$, and $C P$ intersect $B C, C A$, and $A B$ at points $D, E$, and $F$, respectively. Prove that if two of the quadrilaterals $A B D E, B C E F, C A F D, A E P F, B F P D$, and $C D P E$ are concyclic, then all six are concyclic.
|
Solution. We first prove the following lemma:
Lemma 1. Let $A B C D$ be a convex quadrilateral and let $A B \cap C D=E$ and $B C \cap D A=F$. Then the circumcircles of triangles $A B F, C D F, B C E$ and $D A E$ all pass through a common point $P$. This point lies on line $E F$ if and only if $A B C D$ in concyclic.
Proof. Let the circumcircles of $A B F$ and $B C F$ intersect at $P \neq B$. We have
$$
\begin{aligned}
\Varangle F P C & =\Varangle F P B+\Varangle B P C=\Varangle B A D+\Varangle B E C=\Varangle E A D+\Varangle A E D= \\
& =180^{\circ}-\Varangle A D E=180^{\circ}-\Varangle F D C
\end{aligned}
$$
which gives us $F, P, C$ and $D$ are concyclic. Similarly we have
$$
\begin{aligned}
\Varangle A P E & =\Varangle A P B+\Varangle B P E=\Varangle A F B+\Varangle B C D=\Varangle D F C+\Varangle F C D= \\
& =180^{\circ}-\Varangle F D C=180^{\circ}-\Varangle A D E
\end{aligned}
$$
which gives us $E, P, A$ and $D$ are concyclic. Since $\Varangle F P E=\Varangle F P B+\Varangle E P B=\Varangle B A D+$ $\Varangle B C D$ we get that $\Varangle F P E=180^{\circ}$ if and only if $\Varangle B A D+\Varangle B C D=180^{\circ}$ which completes the lemma. We now divide the problem into cases:
Case 1: $A E P F$ and $B F E C$ are concyclic. Here we get that
$$
180^{\circ}=\Varangle A E P+\Varangle A F P=360^{\circ}-\Varangle C E B-\Varangle B F C=360^{\circ}-2 \Varangle C E B
$$
and here we get that $\Varangle C E B=\Varangle C F B=90^{\circ}$, from here it follows that $P$ is the ortocenter of $\triangle A B C$ and that gives us $\Varangle A D B=\Varangle A D C=90^{\circ}$. Now the quadrilaterals $C E P D$ and $B D P F$ are concyclic because
$$
\Varangle C E P=\Varangle C D P=\Varangle P D B=\Varangle P F B=90^{\circ} .
$$
Quadrilaterals $A C D F$ and $A B D E$ are concyclic because
$$
\Varangle A E B=\Varangle A D B=\Varangle A D C=\Varangle A F C=90^{\circ}
$$
Case 2: $A E P F$ and $C E P D$ are concyclic. Now by lemma 1 applied to the quadrilateral $A E P F$ we get that the circumcircles of $C E P, C A F, B P F$ and $B E A$ intersect at a point on $B C$. Since $D \in B C$ and $C E P D$ is concyclic we get that $D$ is the desired point and it follows that $B D P F, B A E D, C A F D$ are all concylic and now we can finish same as Case 1 since $A E D B$ and $C E P D$ are concyclic.
Case 3: $A E P F$ and $A E D B$ are concyclic. We apply lemma 1 as in Case 2 on the quadrilateral $A E P F$. From the lemma we get that $B D P F, C E P D$ and $C A F D$ are concylic and we finish off the same as in Case 1.
Case 4: $A C D F$ and $A B D E$ are concyclic. We apply lemma 1 on the quadrilateral $A E P F$ and get that the circumcircles of $A C F, E C P, P F B$ and $B A E$ intersect at one point. Since this point is $D$ (because $A C D F$ and $A B D E$ are concyclic) we get that $A E P F, C E P D$ and $B F P D$ are concylic. We now finish off as in Case 1. These four cases prove the problem statement.
Remark. A more natural approach is to solve each of the four cases by simple angle chasing.
## Number Theory
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A2. Let $a, b, c$ be positive real numbers such that abc $=1$. Show that
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{(a b+b c+c a)^{2}}{6}
$$
so
|
Solution. By the AM-GM inequality we have $a^{3}+b c \geq 2 \sqrt{a^{3} b c}=2 \sqrt{a^{2}(a b c)}=2 a$ and
$$
\frac{1}{a^{3}+b c} \leq \frac{1}{2 a}
$$
Similarly; $\frac{1}{b^{3}+c a} \leq \frac{1}{2 b} \cdot \frac{1}{c^{3}+a b} \leq \frac{1}{2 c}$ and then
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{1}{2 a}+\frac{1}{2 b}+\frac{1}{2 c}=\frac{1}{2} \frac{a b+b c+c a}{a b c} \leq \frac{(a b+b c+c a)^{2}}{6}
$$
Therefore it is enongil to prove $\frac{(a h+b c+c a)^{2}}{6} \leq \frac{(a b+b c+c a)^{2}}{6}$. This mequalits is trivially shomn to be equivalent to $3 \leq a b+b c+c a$ which is true because of the AM-GM inequalit: $3=\sqrt[3]{(a b c)^{2}} \leq a b+b c+c a$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A3. Let $a . b$ c ce positue real numbers such that $a+b+c=a^{2}+b^{2}+c^{2}$. Shou that
$$
\frac{a^{2}}{a^{2}+a b}+\frac{b^{2}}{b^{2}+b c}+\frac{c^{2}}{c^{2}+c a} \geq \frac{a+b+c}{2}
$$
|
Solution. By the Cauchy-Schwarz inequality it is
$$
\begin{aligned}
& \left(\frac{a^{2}}{a^{2}+a b}+\frac{b^{2}}{b^{2}+b c}+\frac{c^{2}}{c^{2}+c a}\right)\left(\left(a^{2}+a b\right)+\left(b^{2}+b c\right)+\left(c^{2}+c a\right)\right) \geq(a+b+c)^{2} \\
\Rightarrow & \frac{a^{2}}{a^{2}+a b}+\frac{b^{2}}{b^{2}+b c}+\frac{c^{2}}{c^{2}+c a} \geq \frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+a b+b c+c a}
\end{aligned}
$$
So in is enough to prove $\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+a b+b c+c a} \geq \frac{a+b+c}{2}$, that is to prove
$$
2(a+b+c) \geq a^{2}+b^{2}+c^{2}+a b+b c+c a
$$
Substituting $a^{2}+b^{2}+c^{2}$ for $a+b+c$ into the left hand side we wish equivalently to prove
$$
a^{2}+b^{2}+c^{2} \geq a b+b c+c a
$$
But the $a^{2}+b^{2} \geq 2 a b, b^{2}+c^{2} \geq 2 b c, c^{2}+a^{2} \geq 2 c a$ which by addition imply the desired inequality.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A3. Show that
$$
\left(a+2 b+\frac{2}{a+1}\right)\left(b+2 a+\frac{2}{b+1}\right) \geq 16
$$
for all positive real numbers $a, b$ satisfying $a b \geq 1$.
|
Solution 1. By the AM-GM Inequality we have:
$$
\frac{a+1}{2}+\frac{2}{a+1} \geq 2
$$
Therefore
$$
a+2 b+\frac{2}{a+1} \geq \frac{a+3}{2}+2 b
$$
and, similarly,
$$
b+2 a+\frac{2}{b+1} \geq 2 a+\frac{b+3}{2}
$$
On the other hand,
$$
(a+4 b+3)(b+4 a+3) \geq(\sqrt{a b}+4 \sqrt{a b}+3)^{2} \geq 64
$$
by the Cauchy-Schwarz Inequality as $a b \geq 1$, and we are done.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A3 MNE
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2
$$
|
## Solution:
Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
$$
\begin{aligned}
2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\right)+2 \sqrt[3]{\frac{c}{a}} \\
& \geq \frac{a}{b}+3 \sqrt[3]{\frac{a}{b}} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} \\
& =\frac{a}{b}+3 \sqrt[3]{\frac{a}{c}}+2 \sqrt[3]{\frac{c}{a}} \\
& \left.=\frac{a}{b}+\sqrt[3]{\frac{a}{c}}+2 \sqrt[3]{\frac{a}{c}}+\sqrt[3]{\frac{c}{a}}\right) \\
& \geq \frac{a}{b}+\sqrt[3]{\frac{a}{c}}+2 \cdot 2 \sqrt{\sqrt[3]{\frac{a}{c}} \sqrt[3]{\frac{c}{a}}} \\
& =\frac{a}{b}+\sqrt[3]{\frac{a}{c}}+4 \\
& >4
\end{aligned}
$$
which yields the given inequality.
(A4) GRE
Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
|
4
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A5 MKCD
Let $x, y, z$ be positive real numbers that satisfy the equality $x^{2}+y^{2}+z^{2}=3$. Prove that
$$
\frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2
$$
|
## Solution:
We have
$$
\begin{aligned}
& \frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2 \Leftrightarrow \\
& \frac{x^{2}+y z+1}{x^{2}+y z+1}+\frac{y^{2}+z x+1}{y^{2}+z x+1}+\frac{z^{2}+x y+1}{z^{2}+x y+1} \leq 2+\frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \Leftrightarrow \\
& 3 \leq 2+\frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \Leftrightarrow \\
& 1 \leq \frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \\
& \frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^{2}+x y+1} \geq \frac{9}{x^{2}+y z+1+y^{2}+z x+1+z^{2}+x y+1}= \\
& \frac{9}{x^{2}+y^{2}+z^{2}+x y+y z+z x+3} \geq \frac{9}{2 x^{2}+y^{2}+z^{2}+3}=1
\end{aligned}
$$
The first inequality: AM-GM inequality (also can be achieved with Cauchy-BunjakowskiSchwarz inequality). The second inequality: $x y+y z+z x £ x^{2}+y^{2}+z^{2}$ (there are more ways to prove it, AM-GM, full squares etc.)
## GEOMETRY
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT2 BUL
A positive integer is called a repunit, if it is written only by ones. The repunit with $n$ digits will be denoted by $\underbrace{11 \ldots 1}_{n}$. Prove that:
a) the repunit $\underbrace{11 \ldots 1}_{n}$ is divisible by 37 if and only if $n$ is divisible by 3 ;
b) there exists a positive integer $k$ such that the repunit $\underbrace{11 \ldots 1}_{n}$ is divisible by 41 if and only if $n$ is divisible by $k$.
|
## Solution:
a) Let $n=3 m+r$, where $m$ and $r$ are non-negative integers and $r<3$.
Denote by $\underbrace{00 \ldots 0}_{p}$ a recording with $p$ zeroes and $\underbrace{a b c a b c \ldots a b c}_{p x a b c c}$ recording with $p$ times $a b c$. We
have: $\quad \underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_{3 m+r}=\underbrace{11 \ldots 1}_{3 m} \cdot \underbrace{00 \ldots 0}_{r}+\underbrace{11 \ldots 1}_{\tau}=111 \cdot \underbrace{100100 \ldots 100100 \ldots 0}_{(m-1) \times 100}+\underbrace{11 \ldots 1}_{r}$.
Since $111=37.3$, the numbers $\underbrace{11 \ldots 1}_{n}$ and $\underbrace{11 \ldots 1}_{r}$ are equal modulo 37 . On the other hand the numbers 1 and 11 are not divisible by 37 . We conclude that $\underbrace{11 \ldots 1}_{n}$ is divisible by 37 if only if $r=0$, i.e. if and only if $n$ is divisible by 3 .
b) Using the idea from a), we look for a repunit, which is divisible by 41 . Obviously, 1 and 11 are not divisible by 41 , while the residues of 111 and 1111 are 29 and 4 , respectively. We have $11111=41 \cdot 271$. Since 11111 is a repunit with 5 digits, it follows in the same way as in a) that $\underbrace{1}_{11 \ldots 1}$ is divisible by 41 if and only if $n$ is divisible by 5 .
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## NT3 ALB
a) Show that the product of all differences of possible couples of six given positive integers is divisible by 960 (original from Albania).
b) Show that the product of all differences of possible couples of six given positive integers. is divisible by 34560 (modified by problem selecting committee).
|
## Solution:
a) Since we have six numbers then at least two of them have a same residue when divided by 3 , so at least one of the differences in our product is divisible by 3 .
Since we have six numbers then at least two of them have a same residuc when divided by 5 , so at least one of the differences in our product is divisible by 5 .
We may have:
a) six numbers with the same parity
b) five numbers with the same parity
c) four numbers with the same parity
d) three numbers with the same parity
There are $C_{6}^{2}=15$ different pairs, so there are 15 different differences in this product.
a) The six numbers have the same parity; then each difference is divisible by 2 , therefore our product is divisible by $2^{15}$.
b) If we have five numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these five numbers, and the second will be the sixth one. Then $C_{5}^{1}=15=5$ differences are odd. So $15-5=10$ differences are even, so product is divisible by $2^{10}$.
c) If we have four numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these four numbers, and the second will be from the two others numbers. Then $2 \cdot C_{4}^{1}=8$ differences are odd. So $15-8=7$ differences are even, so our product is divisible by $2^{7}$.
d) If we have three numbers with the same parity, then the couples that have their difference odd are formed by taking one from each triple. Then $C_{3}^{1} \cdot C_{3}^{1}=9$ differences are odd, therefore $15-9=6$ differences are even, so our product is divisible by $2^{6}$.
Thus, our production is divisible by $2^{6} \cdot 3 \cdot 5=960$.
b) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be these numbers. Since we have six numbers then at least two of them when divided by 5 have the same residue, so at least one of these differences in our product is divisible by 5 .
Since we have six numbers, and we have three possible residues at the division by 3 , then at least three of them replies the residue of previous numbers, so at least three of these differences in our product are divisible by 3 .
Since we have six numbers, and we have two possible residues at the division by 2 , then at least four of them replies the residue of previous numbers, and two of them replies replied residues, so at least six of these differences in our product are divisible by 2 .
Since we have six numbers, and we have four possible residues at the division by 4 , then at least two of them replies the residue of previous numbers, so at least two of these differences in our product are divisible by 4 . That means that two of these differences are divisible by 4 and moreover four of them are divisible by 2 .
Thus, our production is divisible by $2^{4} \cdot 4^{2} \cdot 3^{3} \cdot 5=34560$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A1 Let $a$ be a real positive number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+a x+a^{2}-6=0$ has no solution in the set of the real number.
|
## Solution
The discriminant of the equation is $\Delta=3\left(8-a^{2}\right)$. If we accept that $\Delta \geq 0$, then $a \leq 2 \sqrt{2}$ and $\frac{1}{a} \geq \frac{\sqrt{2}}{4}$, from where $a^{2} \geq 6+6 \cdot \frac{\sqrt{2}}{4}=6+\frac{6}{a} \geq 6+\frac{3 \sqrt{2}}{2}>8$ (contradiction).
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A2 Prove that $\frac{a^{2}-b c}{2 a^{2}+b c}+\frac{b^{2}-c a}{2 b^{2}+c a}+\frac{c^{2}-a b}{2 c^{2}+a b} \leq 0$ for any real positive numbers $a, b, c$.
|
## Solution
The inequality rewrites as $\sum \frac{2 a^{2}+b c-3 b c}{2 a^{2}+b c} \leq 0$, or $3-3 \sum \frac{b c}{2 a^{2}+b c} \leq 0$ in other words $\sum \frac{b c}{2 a^{2}+b c} \geq 1$.
Using Cauchy-Schwarz inequality we have
$$
\sum \frac{b c}{2 a^{2}+b c}=\sum \frac{b^{2} c^{2}}{2 a^{2} b c+b^{2} c^{2}} \geq \frac{\left(\sum b c\right)^{2}}{2 a b c(a+b+c)+\sum b^{2} c^{2}}=1
$$
as claimed.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A3 Let $A$ be a set of positive integers containing the number 1 and at least one more element. Given that for any two different elements $m, n$ of $A$ the number $\frac{m+1}{(m+1, n+1)}$ is also an element of $A$, prove that $A$ coincides with the set of positive integers.
|
## Solution
Let $a>1$ be lowest number in $A \backslash\{1\}$. For $m=a, n=1$ one gets $y=\frac{a+1}{(2, a+1)} \in A$. Since $(2, a+1)$ is either 1 or 2 , then $y=a+1$ or $y=\frac{a+1}{2}$.
But $1<\frac{a+1}{2}<a$, hence $y=a+1$. Applying the given property for $m=a+1, n=a$ one has $\frac{a+2}{(a+2, a+1)}=a+2 \in A$, and inductively $t \in A$ for all integers $t \geq a$.
Furthermore, take $m=2 a-1, n=3 a-1$ (now in $A!$ ); as $(m+1, n+1)=(2 a, 3 a)=a$ one obtains $\frac{2 a}{a}=2 \in A$, so $a=2$, by the definition of $a$.
The conclusion follows immediately.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A4 Let $a$ and $b$ be positive integers bigger than 2. Prove that there exists a positive integer $k$ and a sequence $n_{1}, n_{2}, \ldots, n_{k}$ consisting of positive integers, such that $n_{1}=a$, $n_{k}=b$, and $\left(n_{i}+n_{i+1}\right) \mid n_{i} n_{i+1}$ for all $i=1,2, \ldots, k-1$.
|
## Solution
We write $a \Leftrightarrow b$ if the required sequence exists. It is clear that $\Leftrightarrow$ is equivalence relation, i.e. $a \Leftrightarrow a,(a \Leftrightarrow b$ implies $b \Rightarrow a)$ and $(a \Leftrightarrow b, b \Leftrightarrow c$ imply $a \Leftrightarrow c$ ).
We shall prove that for every $a \geq 3$, ( $a-$ an integer), $a \Leftrightarrow 3$.
If $a=2^{s} t$, where $t>1$ is an odd number, we take the sequence
$$
2^{s} t, 2^{s}\left(t^{2}-t\right), 2^{s}\left(t^{2}+t\right), 2^{s}(t+1)=2^{s+1} \cdot \frac{t+1}{2}
$$
Since $\frac{t+1}{2}1$ we have $2^{s}, 3 \cdot 2^{s}, 3 \cdot 2^{s-1}, 3 \cdot 2^{s-2}, \ldots, 3$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A5 The real numbers $x, y, z, m, n$ are positive, such that $m+n \geq 2$. Prove that
$$
\begin{gathered}
x \sqrt{y z(x+m y)(x+n z)}+y \sqrt{x z(y+m x)(y+n z)}+z \sqrt{x y(z+m x)(x+n y)} \leq \\
\frac{3(m+n)}{8}(x+y)(y+z)(z+x) .
\end{gathered}
$$
|
## Solution
Using the AM-GM inequality we have
$$
\begin{aligned}
& \sqrt{y z(x+m y)(x+n z)}=\sqrt{(x z+m y z)(x y+n y z)} \leq \frac{x y+x z+(m+n) y z}{2} \\
& \sqrt{x z(y+m x)(y+n z)}=\sqrt{(y z+m x z)(x y+n x z)} \leq \frac{x y+y z+(m+n) x z}{2} \\
& \sqrt{x y(z+m x)(z+n y)}=\sqrt{(y z+m x y)(x z+n x y)} \leq \frac{x z+y z+(m+n) x y}{2}
\end{aligned}
$$
Thus it is enough to prove that
$$
\begin{aligned}
x[x y+x z+(m+n) y z] & +y[x y+y z+(m+n) x z]+z[x y+y z+(m+n) x z] \leq \\
\leq & \frac{3(m+n)}{4}(x+y)(y+z)(z+x),
\end{aligned}
$$
or
$$
4[A+3(m+n) B] \leq 3(m+n)(A+2 B) \Leftrightarrow 6(m+n) B \leq[3(m+n)-4] A
$$
where $A=x^{2} y+x^{2} z+x y^{2}+y^{2} z+x z^{2}+y z^{2}, B=x y z$.
Because $m+n \geq 2$ we obtain the inequality $m+n \leq 3(m+n)-4$. From AMGM inequality it follows that $6 B \leq A$. From the last two inequalities we deduce that $6(m+n) B \leq[3(m+n)-4] A$. The inequality is proved.
Equality holds when $m=n=1$ and $x=y=z$.
### 2.2 Combinatorics
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C2 Consider 50 points in the plane, no three of them belonging to the same line. The points have been colored into four colors. Prove that there are at least 130 scalene triangles whose vertices are colored in the same color.
|
## Solution
Since $50=4 \cdot 12+2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. We start with the:
Lemma. Given $n>8$ points in the plane, no three of them collinear, then there are at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles with vertices among the given points.
Proof. There are $\frac{n(n-1)}{2}$ segments and $\frac{n(n-1)(n-2)}{6}$ triangles with vertices among the given points. We shall prove that there are at most $n(n-1)$ isosceles triangles. Indeed, for every segment $A B$ we can construct at most two isosceles triangles (if we have three $A B C, A B D$ and $A B E$, than $C, D, E$ will be collinear). Hence we have at least
$$
\frac{n(n-1)(n-2)}{6}-n(n-1)=\frac{n(n-1)(n-8)}{6} \text { scalene triangles. }
$$
For $n=13$ we have $\frac{13 \cdot 12 \cdot 5}{6}=130$, QED.
|
130
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT2 Prove that the equation $x^{2006}-4 y^{2006}-2006=4 y^{2007}+2007 y$ has no solution in the set of the positive integers.
|
## Solution
We assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have
$$
\begin{gathered}
x^{2006}=4 y^{2007}+4 y^{2006}+2007 y+2006 \\
x^{2006}+1=4 y^{2006}(y+1)+2007(y+1) \\
x^{2006}+1=\left(4 y^{2006}+2007\right)(y+1)
\end{gathered}
$$
But $4 y^{2006}+2007 \equiv 3(\bmod 4)$, so $x^{2006}+1$ will have at least one prime divisor of the type $4 k+3$. It is known (and easily obtainable by using Fermat's Little Theorem) that this is impossible.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT3 Let $n>1$ be a positive integer and $p$ a prime number such that $n \mid(p-1)$ and $p \mid\left(n^{6}-1\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square.
|
## Solution
Since $n \mid p-1$, then $p=1+n a$, where $a \geq 1$ is an integer. From the condition $p \mid n^{6}-1$, it follows that $p|n-1, p| n+1, p \mid n^{2}+n+1$ or $p \mid n^{2}-n+1$.
- Let $p \mid n-1$. Then $n \geq p+1>n$ which is impossible.
- Let $p \mid n+1$. Then $n+1 \geq p=1+n a$ which is possible only when $a=1$ and $p=n+1$, i.e. $p-n=1=1^{2}$.
- Let $p \mid n^{2}+n+1$, i.e. $n^{2}+n+1=p b$, where $b \geq 1$ is an integer.
The equality $p=1+n a$ implies $n \mid b-1$, from where $b=1+n c, c \geq 0$ is an integer. We have
$$
n^{2}+n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \text { or } n+1=a c n+a+c
$$
If $a c \geq 1$ then $a+c \geq 2$, which is impossible. If $a c=0$ then $c=0$ and $a=n+1$. Thus we obtain $p=n^{2}+n+1$ from where $p+n=n^{2}+2 n+1=(n+1)^{2}$.
- Let $p \mid n^{2}-n+1$, i.e. $n^{2}-n+1=p b$ and analogously $b=1+n c$. So
$$
n^{2}-n+1=p b=(1+n a)(1+n c)=1+(a+c) n+a c n^{2} \text { or } n-1=a c n+a+c
$$
Similarly, we have $c=0, a=n-1$ and $p=n^{2}-n+1$ from where $p-n=n^{2}-2 n+1=$ $(n-1)^{2}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT4 Let $a, b$ be two co-prime positive integers. A number is called good if it can be written in the form $a x+b y$ for non-negative integers $x, y$. Define the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ as $f(n)=n-n_{a}-n_{b}$, where $s_{t}$ represents the remainder of $s$ upon division by $t$. Show that an integer $n$ is good if and only if the infinite sequence $n, f(n), f(f(n)), \ldots$ contains only non-negative integers.
|
## Solution
If $n$ is good then $n=a x+b y$ also $n_{a}=(b y)_{a}$ and $n_{b}=(a x)_{b}$ so
$$
f(n)=a x-(a x)_{b}+b y-(b y)_{a}=b y^{\prime}+a x^{\prime}
$$
is also good, thus the sequence contains only good numbers which are non-negative.
Now we have to prove that if the sequence contains only non-negative integers then $n$ is good. Because the sequence is non-increasing then the sequence will become constant from some point onwards. But $f(k)=k$ implies that $k$ is a multiple of $a b$ thus some term of the sequence is good. We are done if we prove the following:
Lemma: $f(n)$ is good implies $n$ is good.
Proof of Lemma: $n=2 n-n_{a}-n_{b}-f(n)=a x^{\prime}+b y^{\prime}-a x-b y=a\left(x^{\prime}-x\right)+b\left(y^{\prime}-y\right)$ and $x^{\prime} \geq x$ because $n \geq f(n) \Rightarrow n-n_{a} \geq f(n)-f(n)_{a} \Rightarrow a x^{\prime} \geq a x+b y-(b y)_{a} \geq a x$. Similarly $y^{\prime} \geq y$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT5 Let $p$ be a prime number. Show that $7 p+3^{p}-4$ is not a perfect square.
|
## Solution
Assume that for a prime number $p$ greater than $3, m=7 p+3^{p}-4$ is a perfect square. Let $m=n^{2}$ for some $n \in \mathbb{Z}$. By Fermat's Little Theorem,
$$
m=7 p+3^{p}-4 \equiv 3-4 \equiv-1 \quad(\bmod p)
$$
If $p=4 k+3, k \in \mathbb{Z}$, then again by Fermat's Little Theorem
$$
-1 \equiv m^{2 k+1} \equiv n^{4 k+2} \equiv n^{p-1} \equiv 1 \quad(\bmod p), \text { but } p>3
$$
a contradiction. So $p \equiv 1(\bmod 4)$.
Therefore $m=7 p+3^{p}-4 \equiv 3-1 \equiv 2(\bmod 4)$. But this is a contradiction since 2 is not perfect square in $(\bmod 4)$. For $p=2$ we have $m=19$ and for $p=3$ we have $m=44$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A2
Let $a, b$ and $c$ be positive real numbers such that abc $=\frac{1}{8}$. Prove the inequality
$$
a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq \frac{15}{16}
$$
When does equality hold?
|
Solution1. By using The Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}= \\
& \quad=\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{c^{2}}{4}+\frac{c^{2}}{4}+\frac{c^{2}}{4}+\frac{c^{2}}{4}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq \\
& \quad \geq 1515 \sqrt{\frac{a^{12} b^{12} c^{12}}{4^{12}}}=15 \sqrt[5]{\left(\frac{a b c}{4}\right)^{4}}=15 \sqrt[5]{\left(\frac{1}{32}\right)^{4}}=\frac{15}{16}
\end{aligned}
$$
as desired. Equality holds if and only if $a=b=c=\frac{1}{2}$.
|
\frac{15}{16}
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A3
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold?
|
Solution1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right) \\
& =\left(a b+1+\frac{a}{c}+a\right)+\left(b c+1+\frac{b}{a}+b\right)+\left(c a+1+\frac{c}{b}+c\right) \\
& =a b+b c+c a+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+3+a+b+c
\end{aligned}
$$
Notice that by AM-GM we have $a b+\frac{b}{a} \geq 2 b, b c+\frac{c}{b} \geq 2 c$, and $c a+\frac{a}{c} \geq 2 a$.
Thus,
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq\left(a b+\frac{b}{a}\right)+\left(b c+\frac{c}{b}\right)+\left(c a+\frac{a}{c}\right)+3+a+b+c \geq 3(a+b+c+1)
$$
The equality holds if and only if $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A4
Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{7+2 b}{1+a}+\frac{7+2 c}{1+b}+\frac{7+2 a}{1+c} \geq \frac{69}{4}
$$
When does equality hold?
|
Solution1. The inequality can be written as: $\frac{5+2(1+b)}{1+a}+\frac{5+2(1+c)}{1+b}+\frac{5+2(1+a)}{1+c} \geq \frac{69}{4}$.
We substitute $1+a=x, 1+b=y, 1+c=z$.
So, we have to prove the inequality
$$
\frac{5+2 y}{x}+\frac{5+2 z}{y}+\frac{5+2 x}{z} \geq \frac{69}{4} \Leftrightarrow 5\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right) \geq \frac{69}{4}
$$
where $x, y, z>1$ real numbers and $x+y+z=4$.
We have
- $\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{4}$
- $\frac{y}{x}+\frac{z}{y}+\frac{x}{z} \geq 3 \cdot \sqrt[3]{\frac{y}{x} \cdot \frac{z}{y} \cdot \frac{x}{z}}=3$
Thus, $\frac{5+2 y}{x}+\frac{5+2 z}{y}+\frac{5+2 x}{z}=5\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right) \geq 5 \cdot \frac{9}{4}+2 \cdot 3=\frac{69}{4}$.
The equality holds, when $\left(x=y=z, \frac{y}{x}=\frac{z}{y}=\frac{x}{z}, x+y+z=4\right)$, thus $x=y=z=\frac{4}{3}$, i.e. $a=b=c=\frac{1}{3}$.
## (Egw)
人s
Let $x, y, z$ be non-negative real numbers satisfying $x+y+z=x y z$. Prove that
$$
2\left(x^{2}+y^{2}+z^{2}\right) \geq 3(x+y+z)
$$
and determine when equality occurs.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A6
Let $a, b, c$ be positive real numbers. Prove that
$$
\left(\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2}\right)\left(\left(3 b^{2}+1\right)^{2}+2\left(1+\frac{3}{c}\right)^{2}\right)\left(\left(3 c^{2}+1\right)^{2}+2\left(1+\frac{3}{a}\right)^{2}\right) \geq 48^{3}
$$
When does equality hold?
|
Solution. Let $x$ be a positive real number. By AM-GM we have $\frac{1+x+x+x}{4} \geq x^{\frac{3}{4}}$, or equivalently $1+3 x \geq 4 x^{\frac{3}{4}}$. Using this inequality we obtain:
$$
\left(3 a^{2}+1\right)^{2} \geq 16 a^{3} \text { and } 2\left(1+\frac{3}{b}\right)^{2} \geq 32 b^{-\frac{3}{2}}
$$
Moreover, by inequality of arithmetic and geometric means we have
$$
f(a, b)=\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2} \geq 16 a^{3}+32 b^{-\frac{3}{2}}=16\left(a^{3}+b^{-\frac{3}{2}}+b^{-\frac{3}{2}}\right) \geq 48 \frac{a}{b}
$$
Therefore, we obtain
$$
f(a, b) f(b, c) f(c, a) \geq 48 \cdot \frac{a}{b} \cdot 48 \cdot \frac{b}{c} \cdot 48 \cdot \frac{c}{a}=48^{3}
$$
Equality holds only when $a=b=c=1$.
## 1 IH $^{\text {th J.M. }} 2014$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A8
Let $x, y$ and $z$ be positive real numbers such that $x y z=1$. Prove the inequality
$$
\frac{1}{x(a y+b)}+\frac{1}{y(a z+b)}+\frac{1}{z(a x+b)} \geq 3 \text {, if: }
$$
a) $a=0$ and $b=1$;
b) $a=1$ and $b=0$;
c) $a+b=1$ for $a, b>0$
When does the equality hold true?
|
Solution. a) The inequality reduces to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3$, which follows directly from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
b) Here the inequality reduces to $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3$, i.e. $x+y+z \geq 3$, which also follows from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
c) Let $m, n$ and $p$ be such that $x=\frac{m}{n}, y=\frac{n}{p}$ и $z=\frac{p}{m}$. The inequality reduces to
$$
\frac{n p}{a m n+b m p}+\frac{p m}{a n p+b n m}+\frac{m n}{a p m+b p n} \geq 3
$$
By substituting $u=n p, v=p m$ and $w=m n$, (1) becomes
$$
\frac{u}{a w+b v}+\frac{v}{a u+b w}+\frac{w}{a v+b u} \geq 3
$$
The last inequality is equivalent to
$$
\frac{u^{2}}{a u w+b u v}+\frac{v^{2}}{a u v+b v w}+\frac{w^{2}}{a v w+b u w} \geq 3
$$
Cauchy-Schwarz Inequality implies
$$
\frac{u^{2}}{a u w+b u v}+\frac{v^{2}}{a u v+b v w}+\frac{w^{2}}{a v w+b u w} \geq \frac{(u+v+w)^{2}}{a u w+b u v+a u v+b v w+a v w+b u w}=\frac{(u+v+w)^{2}}{u w+v u+w v}
$$
Thus, the problem simplifies to $(u+v+w)^{2} \geq 3(u w+v u+w v)$, which is equivalent to $(u-v)^{2}+(v-w)^{2}+(w-u)^{2} \geq 0$.
Equality holds only when $u=v=w$, that is only for $x=y=z=1$.
Remark. The problem can be reformulated:
Let $a, b, x, y$ and $z$ be nonnegative real numbers such that $x y z=1$ and $a+b=1$. Prove the inequality
$$
\frac{1}{x(a y+b)}+\frac{1}{y(a z+b)}+\frac{1}{z(a x+b)} \geq 3
$$
When does the equality hold true?
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## A9
Let $n$ be a positive integer, and let $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ be positive real numbers such that $x_{1}+\ldots+x_{n}=y_{1}+\ldots+y_{n}=1$. Show that
$$
\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right| \leq 2-\min _{1 \leq i \leq n} \frac{x_{i}}{y_{i}}-\min _{1 \leq i \leq n} \frac{y_{i}}{x_{i}}
$$
|
Solution. Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\frac{x_{1}}{y_{1}} \leq \ldots \leq \frac{x_{n}}{y_{n}}$. Let $A=\frac{x_{1}}{y_{1}}$ and $B=\frac{x_{n}}{y_{n}}$, and $\mathrm{S}=\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right|$. Our aim is to prove that $S \leq 2-A-\frac{1}{B}$.
First, note that we cannot have $A>1$, since that would imply $x_{i}>y_{i}$ for all $i \leq n$, hence $x_{1}+\ldots+x_{n}>y_{1}+\ldots+y_{n}$. Similarly, we cannot have $B<1$, since that would imply $x_{i}<y_{i}$ for all $i \leq n$, hence $x_{1}+\ldots+x_{n}<y_{1}+\ldots+y_{n}$.
If $n=1$, then $x_{1}=y_{1}=A=B=1$ and $S=0$, hence $S \leq 2-A-\frac{1}{B}$. For $n \geq 2$ let $1 \leq k<n$ be some integer such that $\frac{x_{k}}{y_{k}} \leq 1 \leq \frac{x_{k+1}}{y_{k+1}}$. We define the positive real numbers $X_{1}=x_{1}+\ldots+x_{k}$, $X_{2}=x_{k+1}+\ldots+x_{n}, Y_{1}=y_{1}+\ldots+y_{k}, Y_{2}=y_{k+1}+\ldots+y_{n}$. Note that $Y_{1} \geq X_{1} \geq A Y_{1}$ and $Y_{2} \leq X_{2} \leq B Y_{2}$. Thus, $A \leq \frac{X_{1}}{Y_{1}} \leq 1 \leq \frac{X_{2}}{Y_{2}} \leq B$. In addition, $S=Y_{1}-X_{1}+X_{2}-Y_{2}$.
From $0<X_{2}, Y_{1} \leq 1,0 \leq Y_{1}-X_{1}$ and $0 \leq X_{2}-Y_{2}$, follows
$$
S=Y_{1}-X_{1}+X_{2}-Y_{2}=\frac{Y_{1}-X_{1}}{Y_{1}}+\frac{X_{2}-Y_{2}}{X_{2}}=2-\frac{X_{1}}{Y_{1}}-\frac{Y_{2}}{X_{2}} \leq 2-A-\frac{1}{B}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## C1
Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\frac{a b}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:
a) the length of the last remaining segment does not depend on the order of the deletions.
b) for every positive integer $n$, the initial segments on the board can be chosen with distinct integer lengths, such that the last remaining segment has length $n$.
|
Solution. a) Observe that $\frac{1}{\frac{a b}{a+b}}=\frac{1}{a}+\frac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_{1}, a_{2}, \ldots, a_{n}$, and $c$ is the length of the last remaining segment, then $\frac{1}{c}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}$ , proving a).
b) From a) and the equation $\frac{1}{n}=\frac{1}{2 n}+\frac{1}{3 n}+\frac{1}{6 n}$ it follows that if the lengths of the starting segments are $2 n, 3 n$ and $6 n$, then the length of the last remaining segment is $n$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## G4
Let $A B C$ be a triangle such that $\overline{A B} \neq \overline{A C}$. Let $M$ be a midpoint of $\overline{B C}, H$ the orthocenter of $A B C, O_{1}$ the midpoint of $\overline{A H}$ and $O_{2}$ the circumcenter of $B C H$. Prove that $O_{1} A M O_{2}$ is a parallelogram.
|
Solution1. Let $O_{2}^{\prime}$ be the point such that $O_{1} A M O_{2}^{\prime}$ is a parallelogram. Note that $\overrightarrow{M O_{2}}=\overrightarrow{A O_{1}}=\overrightarrow{O_{1} H}$. Therefore, $O_{1} H O_{2}^{\prime} M$ is a parallelogram and $\overrightarrow{M O_{1}}=\overrightarrow{O_{2} H}$.
Since $M$ is the midpoint of $\overline{B C}$ and $O_{1}$ is the midpoint of $\overline{A H}$, it follows that $4 \overrightarrow{M O_{1}}=\overrightarrow{B A}+\overrightarrow{B H}+\overrightarrow{C A}+\overrightarrow{C H}=2(\overrightarrow{C A}+\overrightarrow{B H})$. Moreover, let $B^{\prime}$ be the midpoint of $\overrightarrow{B H}$. Then,
$$
\begin{aligned}
2 \overrightarrow{O_{2}^{\prime B}} \cdot \overrightarrow{B H} & =\left(\overline{O_{2}^{\prime H}}+\overline{O_{2}^{\prime B}}\right) \cdot \overrightarrow{B H}=\left(2 \overline{O_{2}^{\prime} H}+\overrightarrow{H B}\right) \cdot \overrightarrow{B H}= \\
& =\left(2 \overline{M O_{1}}+\overline{H B}\right) \cdot \overline{B H}=(\overline{C A}+\overrightarrow{B H}+\overrightarrow{H B}) \cdot \overrightarrow{B H}=\overline{C A} \cdot \overrightarrow{B H}=0 .
\end{aligned}
$$
By $\vec{a} \cdot \vec{b}$ we denote the inner product of the vectors $\vec{a}$ and $\vec{b}$.
Therefore, $O_{2}^{\prime}$ lies on the perpendicular bisector of $\overline{B H}$. Since $B$ and $C$ play symmetric roles, $\mathrm{O}_{2}^{\prime}$ also lies on the perpendicular bisector of $\overline{\mathrm{CH}}$, hence $\mathrm{O}_{2}^{\prime}$ is the circumcenter of $\triangle B C H$ and $\mathrm{O}_{2}=\mathrm{O}_{2}^{\prime}$.
Note: The condition $\overline{A B} \neq \overline{A C}$ just aims at ensuring that the parallelogram $O_{1} A N O_{2}$ is not degenerate, hence at helping students to focus on the "general" case.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
N4
Prove there are no integers $a$ and $b$ satisfying the following conditions:
i) $16 a-9 b$ is a prime number
ii) $\quad a b$ is a perfect square
iii) $a+b$ is a perfect square
|
Solution. Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
$$
\begin{aligned}
& 16 a-9 b=p \\
& a b=n^{2} \\
& a+b=m^{2}
\end{aligned}
$$
Moreover, let $d=g d c(a, b)$ and $a=d x, b=d y$ for some relatively prime integers $x$ and $y$. Obviously $a \neq 0$ and $b \neq 0, a$ and $b$ are positive (by (2) and (3)).
From (2) follows that $x$ and $y$ are perfect squares, say $x=l^{2}$ and $y=s^{2}$.
From (1), $d \mid p$ and hence $d=p$ or $d=1$. If $d=p$, then $16 x-9 y=1$, and we obtain $x=9 k+4$, $y=16 k+7$ for some nonnegative integer $k$. But then $s^{2}=y \equiv 3(\bmod 4)$, which is a contradiction.
If $d=1$ then $16 l^{2}-9 s^{2}=p \Rightarrow(4 l-3 s)(4 l+3 s)=p \Rightarrow(4 l+3 s=p \wedge 4 l-3 s=1)$.
By adding the last two equations we get $8 l=p+1$ and by subtracting them we get $6 s=p-1$. Therefore $p=24 t+7$ for some integer $t$ and $a=(3 t+1)^{2}$ and $b=(4 t+1)^{2}$ satisfy the conditions (1) and (2). By (3) we have $m^{2}=(3 t+1)^{2}+(4 t+1)^{2}=25 t^{2}+14 t+2$, or equivalently $25 m^{2}=(25 t+7)^{2}+1$.
Since the difference between two nonzero perfect square cannot be 1 , we have a contradiction. As a result there is no solution.
## NS
Find all nonnegative integers $x, y, z$ such that
$$
2013^{x}+2014^{y}=2015^{z}
$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A2. Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
|
Solution. The given inequality is equivalent to
$$
\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}\right) \geqslant a+b+c
$$
By the AM-GM Inequality it follows that
$$
a^{3}+b^{3}=\frac{a^{3}+a^{3}+b^{3}}{3}+\frac{b^{3}+b^{3}+a^{3}}{3} \geqslant a^{2} b+b^{2} a=a b(a+b)
$$
Similarly we have
$$
b^{3}+c^{3} \geqslant b c(b+c) \quad \text { and } \quad c^{3}+a^{3} \geqslant c a(c+a)
$$
Summing the three inequalities we get
$$
2\left(a^{3}+b^{2}+c^{3}\right) \geqslant(a b(a+b)+b c(b+c)+c a(c+a))
$$
From the Cauchy-Schwarz Inequality we have
$$
(a b(a+b)+b c(b+c)+c a(c+a))\left(\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}\right) \geqslant(a b+b c+c a)^{2}
$$
We also have
$$
(a b+b c+c a)^{2} \geqslant 3(a b \cdot b c+b c \cdot c a+c a \cdot a b)=3 a b c(a+b+c)=2(a+b+c)
$$
Combining together (2),(3) and (4) we obtain (1) which is the required inequality.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A4. Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c
$$
Prove that $-\sqrt{c}<a b<0$.
|
Solution. Firstly, we see that
$$
2019(a-b)=a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)
$$
Since $a \neq b$, we get $(a+b)\left(a^{2}+b^{2}\right)=2019$, so $a+b \neq 0$. Thus
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& =-2 a b\left(a^{2}+a b+b^{2}\right)
\end{aligned}
$$
Hence $a b\left(a^{2}+a b+b^{2}\right)=-c-a b$ (the equality does not occur since $a+b \neq 0)$. So
$$
-c=a b\left(a^{2}+a b+b^{2}\right)<-(a b)^{2} \Longrightarrow(a b)^{2}<c \Rightarrow-\sqrt{c}<a b<\sqrt{c}
$$
Therefore, we have $-\sqrt{c}<a b<0$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A5. Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove the inequality
$$
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d}+\frac{1}{a+b+c+d^{3}} \leqslant \frac{a+b+c+d}{4}
$$
|
Solution. From the Cauchy-Schwarz Inequality, we obtain
$$
(a+b+c+d)^{2} \leqslant\left(a^{3}+b+c+d\right)\left(\frac{1}{a}+b+c+d\right)
$$
Using this, together with the other three analogous inequalities, we get
$$
\begin{aligned}
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d} & +\frac{1}{a+b+c+d^{3}} \\
& \leqslant \frac{3(a+b+c+d)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}{(a+b+c+d)^{2}}
\end{aligned}
$$
So it suffices to prove that
$$
(a+b+c+d)^{3} \geqslant 12(a+b+c+d)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)
$$
or equivalently, that
$$
\begin{aligned}
& \left(a^{3}+b^{3}+c^{3}+d^{3}\right)+3 \sum a^{2} b+6(a b c+a b d+a c d+b c d) \\
& \quad \geqslant 12(a+b+c+d)+4(a b c+a b d+a c d+b c d)
\end{aligned}
$$
(Here, the sum is over all possible $x^{2} y$ with $x, y \in\{a, b, c, d\}$ and $x \neq y$.) From the AM-GM Inequality we have
$a^{3}+a^{2} b+a^{2} b+a^{2} c+a^{2} c+a^{2} d+a^{2} d+b^{2} a+c^{2} a+d^{2} a+b c d+b c d \geqslant 12 \sqrt[12]{a^{18} b^{6} c^{6} d^{6}}=12 a$.
Similarly, we get three more inequalities. Adding them together gives the inequality we wanted. Equality holds if and only if $a=b=c=d=1$.
Remark by PSC. Alternatively, we can finish off the proof by using the following two inequalities: Firstly, we have $a+b+c+d \geqslant 4 \sqrt[4]{a b c d}=4$ by the AM-GM Inequality, giving
$$
\frac{3}{4}(a+b+c+d)^{3} \geqslant 12(a+b+c+d)
$$
Secondly, by Mclaurin's Inequality, we have
$$
\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant \frac{b c d+a c d+a b d+a b c}{4}
$$
giving
$$
\frac{1}{4}(a+b+c+d)^{3} \geqslant 4(b c d+a c d+a b d+a b c)
$$
Adding those inequlities we get the required result.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A6. Let $a, b, c$ be positive real numbers. Prove the inequality
$$
\left(a^{2}+a c+c^{2}\right)\left(\frac{1}{a+b+c}+\frac{1}{a+c}\right)+b^{2}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)>a+b+c
$$
|
Solution. By the Cauchy-Schwarz Inequality, we have
$$
\frac{1}{a+b+c}+\frac{1}{a+c} \geqslant \frac{4}{2 a+b+2 c}
$$
and
$$
\frac{1}{b+c}+\frac{1}{a+b} \geqslant \frac{4}{a+2 b+c}
$$
Since
$$
a^{2}+a c+c^{2}=\frac{3}{4}(a+c)^{2}+\frac{1}{4}(a-c)^{2} \geqslant \frac{3}{4}(a+c)^{2}
$$
then, writing $L$ for the Left Hand Side of the required inequality, we get
$$
L \geqslant \frac{3(a+c)^{2}}{2 a+b+2 c}+\frac{4 b^{2}}{a+2 b+c}
$$
Using again the Cauchy-Schwarz Inequality, we have:
$$
L \geqslant \frac{(\sqrt{3}(a+c)+2 b)^{2}}{3 a+3 b+3 c}>\frac{(\sqrt{3}(a+c)+\sqrt{3} b)^{2}}{3 a+3 b+3 c}=a+b+c
$$
Alternative Question by Proposers. Let $a, b, c$ be positive real numbers. Prove the inequality
$$
\frac{a^{2}}{a+c}+\frac{b^{2}}{b+c}>\frac{a b-c^{2}}{a+b+c}+\frac{a b}{a+b}
$$
Note that both this inequality and the original one are equivalent to
$$
\left(c+\frac{a^{2}}{a+c}\right)+\left(a-\frac{a b-c^{2}}{a+b+c}\right)+\frac{b^{2}}{b+c}+\left(b-\frac{a b}{a+b}\right)>a+b+c
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A7. Show that for any positive real numbers $a, b, c$ such that $a+b+c=a b+b c+c a$, the following inequality holds
$$
3+\sqrt[3]{\frac{a^{3}+1}{2}}+\sqrt[3]{\frac{b^{3}+1}{2}}+\sqrt[3]{\frac{c^{3}+1}{2}} \leqslant 2(a+b+c)
$$
|
Solution. Using the condition we have
$$
a^{2}-a+1=a^{2}-a+1+a b+b c+c a-a-b-c=(c+a-1)(a+b-1)
$$
Hence we have
$$
\sqrt[3]{\frac{a^{3}+1}{2}}=\sqrt[3]{\frac{(a+1)\left(a^{2}-a+1\right)}{2}}=\sqrt[3]{\left(\frac{a+1}{2}\right)(c+a-1)(a+b-1)}
$$
Using the last equality together with the AM-GM Inequality, we have
$$
\begin{aligned}
\sum_{\mathrm{cyc}} \sqrt[3]{\frac{a^{3}+1}{2}} & =\sum_{\mathrm{cyc}} \sqrt[3]{\left(\frac{a+1}{2}\right)(c+a-1)(a+b-1)} \\
& \leqslant \sum_{\mathrm{cyc}} \frac{\frac{a+1}{2}+c+a-1+a+b-1}{3} \\
& =\sum_{c y c} \frac{5 a+2 b+2 c-3}{6} \\
& =\frac{3(a+b+c-1)}{2}
\end{aligned}
$$
Hence it is enough to prove that
$$
3+\frac{3(a+b+c-1)}{2} \leqslant 2(a+b+c)
$$
or equivalently, that $a+b+c \geqslant 3$. From a well- known inequality and the condition, we have
$$
(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3(a+b+c)
$$
thus $a+b+c \geqslant 3$ as desired.
Alternative Proof by PSC. Since $f(x)=\sqrt[3]{x}$ is concave for $x \geqslant 0$, by Jensen's Inequality we have
$$
\sqrt[3]{\frac{a^{3}+1}{2}}+\sqrt[3]{\frac{b^{3}+1}{2}}+\sqrt[3]{\frac{c^{3}+1}{2}} \leqslant 3 \sqrt[3]{\frac{a^{3}+b^{3}+c^{3}+3}{6}}
$$
So it is enough to prove that
$$
\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}+3}{6}} \leqslant \frac{2(a+b+c)-3}{3}
$$
We now write $s=a+b+c=a b+b c+c a$ and $p=a b c$. We have
$$
a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)=s^{2}-2 s
$$
and
$$
r=a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}=(a b+b c+c a)(a+b+c)-3 a b c=s^{2}-3 p
$$
Thus,
$$
a^{3}+b^{3}+c^{3}=(a+b+c)^{3}-3 r-6 a b c=s^{3}-3 s^{2}+3 p
$$
So to prove (1), it is enough to show that
$$
\frac{s^{3}-3 s^{2}+3 p+3}{6} \leqslant \frac{(2 s-3)^{3}}{27}
$$
Expanding, this is equivalent to
$$
7 s^{3}-45 s^{2}+108 s-27 p-81 \geqslant 0
$$
By the AM-GM Inequality we have $s^{3} \geqslant 27 p$. So it is enough to prove that $p(s) \geqslant 0$, where
$$
p(s)=6 s^{3}-45 s^{2}+108 s-81=3(s-3)^{2}(2 s-3)
$$
It is easy to show that $s \geqslant 3$ (e.g. as in the first solution) so $p(s) \geqslant 0$ as required.
## COMBINATORICS
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C1. Let $S$ be a set of 100 positive integers having the following property:
"Among every four numbers of $S$, there is a number which divides each of the other three or there is a number which is equal to the sum of the other three."
Prove that the set $S$ contains a number which divides each of the other 99 numbers of $S$.
|
Solution. Let $a<b$ be the two smallest numbers of $S$ and let $d$ be the largest number of $S$. Consider any two other numbers $x<y$ of $S$. For the quadruples $(a, b, x, d)$ and $(a, b, y, d)$ we cannot get both of $d=a+b+x$ and $d=a+b+y$, since $a+b+x<a+b+y$. From here, we get $a \mid b$ and $a \mid d$.
Consider any number $s$ of $S$ different from $a, b, d$. From the condition of the problem, we get $d=a+b+s$ or $a$ divides $b, s$ and $d$. But since we already know that $a$ divides $b$ and $d$ anyway, we also get that $a \mid s$, as in the first case we have $s=d-a-b$. This means that $a$ divides all other numbers of $S$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C2. In a certain city there are $n$ straight streets, such that every two streets intersect, and no three streets pass through the same intersection. The City Council wants to organize the city by designating the main and the side street on every intersection. Prove that this can be done in such way that if one goes along one of the streets, from its beginning to its end, the intersections where this street is the main street, and the ones where it is not, will apear in alternating order.
|
Solution. Pick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.
On every other street $s_{1}$, exactly one intersection has been organized, namely the one where $s_{1}$ intersects $s$. Call this intersection $I_{1}$. We want to organize the intersections along $s_{1}$ such that they alternate between the two types. Note that, as $I_{1}$ is already organized, we have exactly one way to organize the remaining intersections along $s_{1}$.
For every street $s_{1} \neq s$, we can apply the procedure described above. Now, we only need to show that every intersection not on $s$ is well-organized. More precisely, this means that for every two streets $s_{1}, s_{2} \neq s$ intersecting at $s_{1} \cap s_{2}=A, s_{1}$ is the main street on $A$ if and only if $s_{2}$ is the side street on $A$.
Consider also the intersections $I_{1}=s_{1} \cap s$ and $I_{2}=s_{2} \cap s$. Now, we will define the "role" of the street $t$ at the intersection $X$ as "main" if this street $t$ is the main street on $X$, and "side" otherwise. We will prove that the roles of $s_{1}$ and $s_{2}$ at $A$ are different.
Consider the path $A \rightarrow I_{1} \rightarrow I_{2} \rightarrow A$. Let the number of intersections between $A$ and $I_{1}$ be $u_{1}$, the number of these between $A$ and $I_{2}$ be $u_{2}$, and the number of these between $I_{1}$ and $I_{2}$ be $v$. Now, if we go from $A$ to $I_{1}$, we will change our role $u_{1}+1$ times, as we will encounter $u_{1}+1$ new intersections. Then, we will change our street from $s_{1}$ to $s$, changing our role once more. Then, on the segment $I_{1} \rightarrow I_{2}$, we have $v+1$ new role changes, and after that one more when we change our street from $s_{1}$ to $s_{2}$. The journey from $I_{2}$ to $A$ will induce $u_{2}+1$ new role changes, so in total we have changed our role $u_{1}+1+1+v+1+1+u_{2}+1=u_{1}+v+u_{2}+5$, As we try to show that roles of $s_{1}$ and $s_{2}$ differ, we need to show that the number of role changes is odd, i.e. that $u_{1}+v+u_{2}+5$ is odd.
Obviously, this claim is equivalent to $2 \mid u_{1}+v+u_{2}$. But $u_{1}, v$ and $u_{2}$ count the number of intersections of the triangle $A I_{1} I_{2}$ with streets other than $s, s_{1}, s_{2}$. Since every street other than $s, s_{1}, s_{2}$ intersects the sides of $A I_{1} I_{2}$ in exactly two points, the total number of intersections is even. As a consequence, $2 \mid u_{1}+v+u_{2}$ as required.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT1. Let $a, b, p, q$ be positive integers such that $a$ and $b$ are relatively prime, $a b$ is even and $p, q \geq 3$. Prove that
$$
2 a^{p} b-2 a b^{q}
$$
$$
a \text { is even }
$$
cannot be a square of an integer number.
|
Solution. Without loss of
Let $a=2 a^{\prime}$. If $\vdots$. Without loss of generality, assume that $a$ is even and consequently $b$ is odd.
$$
2 a^{p} b-2 a b^{q}=4 a^{\prime} b\left(a^{p-1}-b^{q-1}\right)
$$
is a square, then $a^{\prime}, b$ and $a^{p-1}-b^{q-1}$ are pairwise coprime.
On the other hand, $a^{p-1}$ is divisible by 4 and $b^{q-1}$ gives the remainder 1 when divided by 4. It follows that $a^{p-1}-b^{q-1}$ has the form $4 k+3$, a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT4. If the positive integers $x$ and $y$ are such that both $3 x+4 y$ and $4 x+3 y$ are perfect squares, prove that both $x$ and $y$ are multiples of 7 .
|
Solution. Let
$$
3 x+4 y=m^{2}, \quad 4 x+3 y=n^{2}
$$
Then
$$
7(x+y)=m^{2}+n^{2} \Rightarrow 7 \mid m^{2}+n^{2}
$$
Considering $m=7 k+r, \quad r \in\{0,1,2,3,4,5,6\}$ we find that $m^{2} \equiv u(\bmod 7), \quad u \in$ $\{0,1,2,4\}$ and similarly $n^{2} \equiv v(\bmod 7), \quad v \in\{0,1,2,4\}$. Therefore we have either $m^{2}+n^{2} \equiv 0 \quad(\bmod 7)$, when $u=v=0$, or $m^{2}+n^{2} \equiv w(\bmod 7), w \in\{1,2,3,4,5,6\}$. However, from (2) we have that $m^{2}+n^{2} \equiv 0(\bmod \tau)$ and hence $u=v=0$ and
$$
m^{2}+n^{2} \equiv 0 \quad\left(\bmod \tau^{2}\right) \Rightarrow 7(x+y) \equiv 0 \quad\left(\bmod \tau^{2}\right)
$$
and cousequently
$$
x+y \equiv 0 \quad(\bmod 7)
$$
Moreover. from (1) we have $x-y=n^{2}-m^{2}$ and $n^{2}-m^{2} \equiv 0\left(\bmod i^{2}\right)$ (since $\left.u=c=0\right)$, so
$$
x-y \equiv 0 \quad(\bmod 7) .
$$
From (3) and (4) we have tliat $x+y=7 k, x-y=7 l$, where $k$ and $l$ are positive incegers. Hence
$$
2 x=7(k+l), 2 y=7(k-l)
$$
where $k+l$ and $k-l$ are positive integers. It follows that $i \mid 2 x$ and $i \mid 2 y$, and finally $\pi \mid x$ and i|y.
## ALGEBRA
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A1. Prove that
$$
(1+a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 3+a+b+c
$$
for any real numbers $a, b, c \geq 1$.
|
Solution. The inequality rewrites as
$$
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+(b c+a c+a b) \geq 3+a+b+c
$$
or
$$
\left(\frac{\frac{1}{a}+\frac{1}{b}}{2}+\frac{\frac{1}{b}+\frac{1}{c}}{2}+\frac{\frac{1}{a}+\frac{1}{c}}{2}\right)+(b c+a c+a b) \geq 3+a+b+c
$$
which is equivalent to
$$
\frac{(2 a b-(a+b))(a b-1)}{2 a b}+\frac{(2 b c-(b+c))(b c-1)}{2 b c}+\frac{(2 c a-(c+a))(c a-1)}{2 c a} \geq 0
$$
The last inequality is true due to the obvious relations
$$
2 x y-(x+y)=x(y-1)+y(x-1) \geq 0
$$
for any two real numbers $x, y \geq 1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A2. Prove that, for all real numbers $x, y, z$ :
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq(x+y+z)^{2}
$$
When the equality holds?
|
Solution. For $x=y=z=0$ the equality is valid.
Since $(x+y+z)^{2} \geq 0$ it is enongh to prove that
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq 0
$$
which is equivalent to the inequality
$$
\frac{x^{2}-y^{2}}{x^{2}+\frac{1}{2}}+\frac{y^{2}-z^{2}}{y^{2}+\frac{1}{2}}+\frac{z^{2}-x^{2}}{z^{2}+\frac{1}{2}} \leq 0
$$
Dellote
$$
a=x^{2}+\frac{1}{2}, b=y^{2}+\frac{1}{2}, c=z^{2}+\frac{1}{2}
$$
Then (1) is equivalent to
$$
\frac{a-b}{a}+\frac{b-c}{b}+\frac{c-a}{c} \leq 0
$$
From very well known $A G$ inequality follows that
$$
a^{2} b+b^{2} c+c^{2} a \geq 3 a b c
$$
Fron the equivalencies
$$
a^{2} b+b^{2} c+c^{2} a \geq 3 a b c \Leftrightarrow \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geq-3 \Leftrightarrow \frac{a-b}{a}+\frac{b-c}{b}+\frac{c-a}{c} \leq 0
$$
follows thait the inequality (2) is valid, for positive real numbers $a, b, c$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A3. Prove that for all real $x, y$
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{2 \sqrt{2}}{\sqrt{x^{2}+y^{2}}}
$$
|
Solution. The inequality rewrites as
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{\sqrt{2\left(x^{2}+y^{2}\right)}}{\frac{x^{2}+y^{2}}{2}}
$$
Now it is enough to prove the next two simple inequalities:
$$
x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}, \quad x^{2}-x y+y^{2} \geq \frac{x^{2}+y^{2}}{2}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A4. Prove that if $0<\frac{a}{b}<b<2 a$ then
$$
\frac{2 a b-a^{2}}{7 a b-3 b^{2}-2 a^{2}}+\frac{2 a b-b^{2}}{7 a b-3 a^{2}-2 b^{2}} \geq 1+\frac{1}{4}\left(\frac{a}{b}-\frac{b}{a}\right)^{2}
$$
|
Solution. If we denote
$$
u=2-\frac{a}{b}, \quad v=2-\frac{b}{a}
$$
then the inequality rewrites as
$$
\begin{aligned}
& \frac{u}{v+u v}+\frac{v}{u+u v} \geq 1+\frac{1}{4}(u-u)^{2} \\
& \frac{(u-v)^{2}+u v(1-u v)}{u v(u v+u+v+1)} \geq \frac{(u-u)^{2}}{4}
\end{aligned}
$$
$\mathrm{Or}$
Since $u>0, v>0, u+v \leq 2, u v \leq 1, u v(u+v+u v+1) \leq 4$, the result is clear.
## GEOMETRY
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
G1. Two circles $k_{1}$ and $k_{2}$ intersect a.t points $A$ and $B$. A circle $k_{3}$ centered at $A$ meet $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $M P$ and $A B>A M$.
Prove rhat the angles $\angle M B Q$ and $\angle N B P$ are equal.
|
Solution As $A M=A P$, we have
$$
\angle M B A=\frac{1}{2} \operatorname{arcAM}=\frac{1}{2} \operatorname{arc} A P=\angle A B P
$$
and likewise
$$
\angle Q B A=\frac{1}{2} \operatorname{arc} A Q=\frac{1}{2} \operatorname{arc} c A N=\angle A B N
$$
Summing these equalities yields $\angle M B Q=\angle N B P$ as needed.
Q2. Let $E$ and $F$ be two distinct points inside of a parallelogram $A B C D$. Find the maximum number of triangles with the same area and having the vertices in three of the following five points: $A, B, C, D, E, F$.
|
notfound
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C1. A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that have in common with the poligon only vertices.
Prove that there are 2 more black triangles than white ones.
|
Solution. Denote by $b, r, w$ the number of black, red white triangles respectively.
It is easy to prove that the polygon is divided into $n-2$ triangles, hence
$$
b+r+w=n-2
$$
Each side of the polygon is a side of exactly one triangle of the decomposition, and thus
$$
2 b+r=n
$$
Subtracting the two relations yields $w=b-2$, as needed. allowed:
|
b-2
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C2. Given $m \times n$ table, each cell signed with "-". The following operations are
(i) to change all the signs in entire row to the opposite, i. e. every "-" to "+", and every "+" to "-";
(ii) to change all the signs in entire column to the opposite, i. e. every "-" to "+" and every "+" to " -".
(a) Prove that if $m=n=100$, using the above operations one can not obtain 2004 signs "t".
(b) If $m=1004$, find the least $n>100$ for which 2004 signs " + " can be obtained.
|
Solution. If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$
(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le
$$
50 l+50 k-1 k=1002
$$
Rewrite the lasc equation as
$$
(50-l)(50-h)=2.500-100.2=1498
$$
Since $1498=2 \cdot 7 \cdot 107$, this equation has no solitions in natural numbers.
(b) Let $n=101$. Then we have
$$
(100-k) l+(101-l) k=2004
$$
OI
$$
100 l+101 k-2 l k=2004
$$
l.e.
$$
101 k=2004-100 l+2 l k \div 2(1002-50 l+l k)
$$
Hence $s=2 t$ and we have $101 t=501-25 l+2 l t$. From here we have
$$
t=\frac{501-25 l}{101-2 l}=4+\frac{97-17 l}{101-2 l}
$$
Since $t$ is natural number and $97-17 l<101-2 l$, this is a contradiction, Hence $n \neq 101$. Let $n=1.02$. Then we have
$$
(100-k) l+(102-l) k=2004
$$
or
$$
100 l+102 k-2 l k=2004
$$
$$
50 l+51 k-l k=1002
$$
Rewrite the last equation as
$$
(51-l)(50-k)=25.50-1002=1.548
$$
Since $145 S=2 \cdot 2 \cdot 3 \cdot 3 \cdot 43$ we have $51-l=36$ and $50-k=43$. From here obtain $l=15$ and $k=7$. Indeed,
$$
(100-\bar{\imath}) \cdot 15+(102-1.5) \cdot \overline{7}=93 \cdot 15+87 \cdot 7=1395+609=2004
$$
Hence, the least $n$ is 102 .
|
102
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem A1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0
$$
implying
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}
$$
It follows that $(a+b)(b+c)(c+a)=0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $0=2+3$, a contradiction). Thus $a+b+c+d \neq 0$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem A4. Let $a, b, c$ be positive real numbers such that $a b c(a+b+c)=3$. Prove the inequality
$$
(a+b)(b+c)(c+a) \geq 8
$$
and determine all cases when equality holds.
|
Solution. We have
$A=(a+b)(b+c)(c+a)=\left(a b+a c+b^{2}+b c\right)(c+a)=(b(a+b+c)+a c)(c+a)$,
so by the given condition
$$
A=\left(\frac{3}{a c}+a c\right)(c+a)=\left(\frac{1}{a c}+\frac{1}{a c}+\frac{1}{a c}+a c\right)(c+a)
$$
Aplying the AM-GM inequality for four and two terms respectively, we get
$$
A \geq 4 \sqrt[4]{\frac{a c}{(a c)^{3}}} \cdot 2 \sqrt{a c}=8
$$
From the last part, it is easy to see that inequality holds when $a=c$ and $\frac{1}{a c}=a c$, i.e. $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
Problem A5. The real positive numbers $x, y, z$ satisfy the relations $x \leq 2$, $y \leq 3, x+y+z=11$. Prove that $\sqrt{x y z} \leq 6$.
|
Solution. For $x=2, y=3$ and $z=6$ the equality holds.
After the substitutions $x=2-u, y=3-v$ with $u \in[0,2), v \in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes
$$
(2-u)(3-v)(6+u+v) \leqslant 36
$$
We shall need the following lemma.
Lemma. If real numbers $a$ and $b$ satisfy the relations $00$ and $y \geq 0$.
The equality in (2) holds if $y=0$. The lemma is proved.
By using the lemma we can write the following inequalities:
$$
\begin{gathered}
\frac{6}{6+u} \geqslant \frac{2-u}{2} \\
\frac{6}{6+v} \geqslant \frac{3-v}{3} \\
\frac{6+u}{6+u+v} \geqslant \frac{6}{6+v}
\end{gathered}
$$
By multiplying the inequalities (3), (4) and (5) we obtain:
$$
\begin{gathered}
\frac{6 \cdot 6 \cdot(6+u)}{(6+u)(6+v)(6+u+v)} \geqslant \frac{6(2-u)(3-v)}{2 \cdot 3(6+v)} \Leftrightarrow \\
(2-u)(3-v)(6+u+v) \leqslant 2 \cdot 3 \cdot 6=36 \Leftrightarrow \quad(1)
\end{gathered}
$$
By virtue of lemma, the equality holds if and only if $u=v=0$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A1. Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 6
$$
|
Solution. We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that
$$
(a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)
$$
so that
$$
\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8
$$
Applying AM-GM, we conclude:
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 3 \cdot \sqrt[3]{\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}}=6
$$
Alternatively, we can write LHS as
$$
\frac{b c(a b+4)}{2(b c+4)}+\frac{a c(b c+4)}{2(a c+4)}+\frac{a b(c a+4)}{2(a b+4)}
$$
and then apply AM-GM.
|
6
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A2. Given positive real numbers $a, b, c$, prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(a+c)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$ and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$. Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b c} \geq \frac{4}{\left(a^{2}+b^{2}\right)(c+1)}
$$
Using the AM-GM inequality, we have
$$
\frac{4}{\left(a^{2}+b^{2}\right)(c+1)}+\frac{a^{2}+b^{2}}{2} \geq 2 \sqrt{\frac{2}{c+1}}=\frac{4}{\sqrt{2(c+1)}}
$$
respectively
$$
\frac{c+3}{8}=\frac{(c+1)+2}{8} \geq \frac{\sqrt{2(c+1)}}{4}
$$
We conclude that
$$
\frac{4}{\left(a^{2}+b^{2}\right)(c+1)}+\frac{a^{2}+b^{2}}{2} \geq \frac{8}{c+3}
$$
and finally
$\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(a+c)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A4. If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:
$$
\frac{x+1}{\sqrt{x^{5}+x+1}}+\frac{y+1}{\sqrt{y^{5}+y+1}}+\frac{z+1}{\sqrt{z^{5}+z+1}} \geq 3
$$
When does the equality hold?
|
Solution. First we factor $x^{5}+x+1$ as follows:
$$
\begin{aligned}
x^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\left(x^{3}-1\right)+x^{2}+x+1=x^{2}(x-1)\left(x^{2}+x+1\right)+x^{2}+x+1 \\
& =\left(x^{2}+x+1\right)\left(x^{2}(x-1)+1\right)=\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)
\end{aligned}
$$
Using the $A M-G M$ inequality, we have
$$
\sqrt{x^{5}+x+1}=\sqrt{\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)} \leq \frac{x^{2}+x+1+x^{3}-x^{2}+1}{2}=\frac{x^{3}+x+2}{2}
$$
and since
$x^{3}+x+2=x^{3}+1+x+1=(x+1)\left(x^{2}-x+1\right)+x+1=(x+1)\left(x^{2}-x+1+1\right)=(x+1)\left(x^{2}-x+2\right)$,
then
$$
\sqrt{x^{5}+x+1} \leq \frac{(x+1)\left(x^{2}-x+2\right)}{2}
$$
Using $x^{2}-x+2=\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}>0$, we obtain $\frac{x+1}{\sqrt{x^{5}+x+1}} \geq \frac{2}{x^{2}-x+2}$ Applying the CauchySchwarz inequality and the given condition, we get
$$
\sum_{c y c} \frac{x+1}{\sqrt{x^{5}+x+1}} \geq \sum_{c y c} \frac{2}{x^{2}-x+2} \geq \frac{18}{\sum_{c y c}\left(x^{2}-x+2\right)}=\frac{18}{6}=3
$$
which is the desired result.
For the equality both conditions: $x^{2}-x+2=y^{2}-y+2=z^{2}-z+2$ (equality in CBS) and $x^{3}-x^{2}+1=x^{2}+x+1$ (equality in AM-GM) have to be satisfied.
By using the given condition it follows that $x^{2}-x+2+y^{2}-y+2+z^{2}-z+2=6$, hence $3\left(x^{2}-x+2\right)=6$, implying $x=0$ or $x=1$. Of these, only $x=0$ satisfies the second condition. We conclude that equality can only hold for $x=y=z=0$.
It is an immediate check that indeed for these values equality holds.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A5. Let $x, y, z$ be positive real numbers such that $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
a) Prove the inequality
$$
x+y+z \geq \sqrt{\frac{x y+1}{2}}+\sqrt{\frac{y z+1}{2}}+\sqrt{\frac{z x+1}{2}}
$$
b) (Added by the problem selecting committee) When does the equality hold?
|
## Solution.
a) We rewrite the inequality as
$$
(\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1})^{2} \leq 2 \cdot(x+y+z)^{2}
$$
and note that, from CBS,
$$
\text { LHS } \leq\left(\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}\right)(x+y+z)
$$
But
$$
\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2(x+y+z)
$$
which proves (1).
b) The equality occurs when we have equality in CBS, i.e. when
$$
\frac{x y+1}{x^{2}}=\frac{y z+1}{y^{2}}=\frac{z x+1}{z^{2}}\left(=\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\right)
$$
Since we can also write
$$
(\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1})^{2} \leq\left(\frac{x y+1}{y}+\frac{y z+1}{z}+\frac{z x+1}{x}\right)(y+z+x)=2(x+y+z)^{2}
$$
the equality implies also
$$
\frac{x y+1}{y^{2}}=\frac{y z+1}{z^{2}}=\frac{z x+1}{x^{2}}\left(=\frac{x y+y z+z x+3}{x^{2}+y^{2}+z^{2}}\right)
$$
But then $x=y=z$, and since $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, we conclude that $x=\frac{1}{x}=1=y=z$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C 2. Viktor and Natalia bought 2020 buckets of ice-cream and want to organize a degustation schedule with 2020 rounds such that:
- In every round, each one of them tries 1 ice-cream, and those 2 ice-creams tried in a single round are different from each other.
- At the end of the 2020 rounds, each one of them has tried each ice-cream exactly once.
We will call a degustation schedule fair if the number of ice-creams that were tried by Viktor before Natalia is equal to the number of ice creams tried by Natalia before Viktor.
Prove that the number of fair schedules is strictly larger than $2020!\left(2^{1010}+(1010!)^{2}\right)$.
|
Solution. If we fix the order in which Natalia tries the ice-creams, we may consider 2 types of fair schedules:
1) Her last 1010 ice-creams get assigned as Viktor's first 1010 ice-creams, and vice versa: Viktor's first 1010 ice-creams are assigned as Natalia's last 1010 ice-creams. This generates (1010!) $)^{2}$ distinct fair schedules by permuting the ice-creams within each group.
2) We divide all ice-creams into disjoint groups of 4 , and in each group we swap the first 2 ice-creams with the last 2 , which gives us $\left((2!)^{2}\right)^{504}=2^{1010}$ distinct schedules.
Now, to make the inequality strict, we consider 1 more schedule like 2 ), but with groups of 2 ice-creams instead of 4 .
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT 5. The positive integer $k$ and the set $A$ of different integers from 1 to $3 k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2 b=a+c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1,2 k]$ - medium and those in $[2 k+1,3 k]$ - large. Is it always true that there are no positive integers $x$ and $d$ such that if $x, x+d$ and $x+2 d$ are divided by $3 k$ then the remainders belong to $A$ and those of $x$ and $x+d$ are different and are:
a) small?
b) medium?
c) large?
(In this problem we assume that if a multiple of $3 k$ is divided by $3 k$ then the remainder is $3 k$ rather than 0. )
|
Solution. A counterexample for a) is $k=3, A=\{1,2,9\}, x=2$ and $d=8$. A counterexample for c) is $k=3, A=\{1,8,9\}, x=8$ and $d=1$.
We will prove that b) is true.
Suppose the contrary and let $x, d$ have the above properties. We can assume $03 k$, then since the remainder for $x+d$ is medium we have $4 k2 k$. Therefore $6 k=4 k+2 kk$ so $d=(x+d)-x<k$. Hence $0 \leq x+2 d=(x+d)+d<3 k$. Thus the remainders $x, x+d$ and $x+2 d$ are in $A$ and
$$
2(x+d)=(x+2 d)+x
$$
a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A 1. Let $x, y$ and $z$ be positive numbers. Prove that
$$
\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{7}}}{\sqrt{2 \sqrt{27}}}
$$
|
Solution. Replacing $x=a^{2}, y=b^{2}, z=c^{2}$, where $a, b, c$ are positive numbers, our inequality is equivalent to
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{\sqrt[4]{(a+b+c)^{7}}}{\sqrt{2 \sqrt{27}}}
$$
Using the Cauchy-Schwarz inequality for the left hand side we get
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{(a+b+c)^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}}}
$$
Using Cauchy-Schwarz inequality for three positive numbers $\alpha . \beta . \uparrow$, we have
$$
\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma} \leq \sqrt{3(\alpha+\beta+\gamma)}
$$
Using this result twice, we have
$$
\begin{aligned}
\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}} & \leq \sqrt{6(\sqrt{a}+\sqrt{b}+\sqrt{c})} \\
& \leq \sqrt{6 \sqrt{3(a+b+c)}}
\end{aligned}
$$
Combining (1) and (2) we get the desired result.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A 3. Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{1}{a b(b+1)(c+1)}+\frac{1}{b c(c+1)(a+1)}+\frac{1}{c a(a+1)(b+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
|
Solution. The required inequality is equivalent to
$$
\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
or equivalently to,
$$
(1+a b c)^{2}(a b+b c+c a+a+b+c) \geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)
$$
Let $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can be rewritten as
$$
(m+n)\left(1+x^{3}\right)^{2} \geq 3 x^{3}\left(x^{3}+m+n+1\right)
$$
or
$$
(m+n)\left(x^{6}-x^{3}+1\right) \geq 3 x^{3}\left(x^{3}+1\right)
$$
By the AM-GM inequality we have $m \geq 3 x$ and $n \geq 3 x^{2}$, hence $m+n \geq 3 x(x+1)$. It is sufficient to prove that
$$
\begin{aligned}
x(x+1)\left(x^{6}-x^{3}+1\right) & \geq x^{3}(x+1)\left(x^{2}-x+1\right) \\
3\left(x^{6}-x^{3}+1\right) & \geq x^{2}\left(x^{2}-x+1\right) \\
\left(x^{2}-1\right)^{2} & \geq 0
\end{aligned}
$$
which is true.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A 5. Let $a, b, c, d$ and $x, y, z, t$ be real numbers such that
$$
0 \leq a, b, c, d \leq 1, \quad x, y, z, t \geq 1 \text { and } a+b+c+d+x+y+z+t=8
$$
Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} \leq 28
$$
When does the equality hold?
|
Solution. We observe that if $u \leq v$ then by replacing $(u, v)$ with $(u-\varepsilon, v+\varepsilon)$, where $\varepsilon>0$, the sum of squares increases. Indeed,
$$
(u-\varepsilon)^{2}+(v+\varepsilon)^{2}-u^{2}-v^{2}=2 \varepsilon(v-u)+2 \varepsilon^{2}>0
$$
Then, denoting
$$
E(a, b, c, d, x, y, z, t)=a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2}
$$
and assuming without loss of generality that $a \leq b \leq c \leq d$ and $x \leq y \leq z \leq t$, we have
$$
\begin{aligned}
E(a, b, c, d, x, y, z, t) & \leq E(0,0,0,0, a+x, b+y, c+z, d+t) \\
& \leq E(0,0,0,0,1, b+y, c+z, a+d+x+t-1) \\
& \leq E(0,0,0,0,1,1, c+z, a+b+d+x+y+t-2) \\
& \leq E(0,0,0,0,1,1,1,5)=28
\end{aligned}
$$
Note that if $(a, b, c, d, x, y, z, t) \neq(0,0,0,0,1,1,1,5)$, at least one of the above inequalities, obtained by the $\epsilon$ replacement mentioned above, should be a strict inequality. Thus, the maximum value of $E$ is 28 , and it is obtained only for $(a, b, c, d, x, y, z, t)=(0,0,0,0,1,1,1,5)$ and permutations of $a, b, c, d$ and of $x, y, z, t$.
|
28
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A 6. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
\frac{a}{\sqrt{a^{3}+5}}+\frac{b}{\sqrt{b^{3}+5}}+\frac{c}{\sqrt{c^{3}+5}} \leq \frac{\sqrt{6}}{2}
$$
|
Solution. From AM-GM inequality we have
$$
a^{3}+a^{3}+1 \geq 3 a^{2} \Rightarrow 2\left(a^{3}+5\right) \geq 3\left(a^{2}+3\right)
$$
Using the condition $a b+b c+c a=3$, we get
$$
\left(a^{3}+5\right) \geq 3\left(a^{2}+a b+b c+c a\right)=3(c+a)(a+b)
$$
therefore
$$
\frac{a}{\sqrt{a^{3}+5}} \leq \sqrt{\frac{2 a^{2}}{3(c+a)(a+b)}}
$$
Using again the AM-GM inequality we get
$$
\sqrt{\frac{2 a^{2}}{3(c+a)(a+b)}} \leq \sqrt{\frac{2}{3}}\left(\frac{\frac{a}{c+a}+\frac{a}{a+b}}{2}\right)=\frac{\sqrt{6}}{6}\left(\frac{a}{c+a}+\frac{a}{a+b}\right)
$$
From (1) and (2) we obtain
$$
\frac{a}{\sqrt{a^{3}+5}} \leq \frac{\sqrt{6}}{6}\left(\frac{a}{c+a}+\frac{a}{a+b}\right)
$$
Similar inequalities hold by cyclic permutations of the $a, b, c$ 's. Adding all these we get
$$
\sum_{\text {cyclic }} \frac{a}{\sqrt{a^{3}+5}} \leq \sum_{\text {cyc }} \frac{\sqrt{6}}{6}\left(\frac{a}{c+a}+\frac{a}{a+b}\right)=\frac{\sqrt{6}}{6} \cdot 3=\frac{\sqrt{6}}{2}
$$
which is the desired result.
|
\frac{\sqrt{6}}{2}
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
G 4. Let $A B C$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be it's incenter. Let $P_{1}, P_{2}$ and $P_{3}$ be the areas of the triangles $A B I, B C I$ and $C A I$, respectively. Prove that
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} \geq 16
$$
|
Solution. Let $r$ be the radius of the inscribed circle of the triangle $A B C$. We have that
$$
P_{1}=\frac{r c}{2}, \quad P_{2}=\frac{r a}{2}, \quad P_{3}=\frac{r b}{2}
$$
It follows that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)
$$
From Leibniz's relation we have that if $H$ is the orthocenter, then
$$
O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2}
$$
It follows that
$$
9 R^{2} \geq a^{2}+b^{2}+c^{2}
$$
Therefore, using the AM-HM inequality and then (1), we get
$$
\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} \geq \frac{9}{a^{2}+b^{2}+c^{2}} \geq \frac{1}{R^{2}}
$$
Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{4}{r^{2} R^{2}} \geq \frac{16}{R^{4}}
$$
Comment by PSC. We can avoid using Leibniz's relation as follows: as in the above solution we have that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)
$$
Let $a+b+c=2 \tau, E=(A B C)$ and using the inequality $x^{2}+y^{2}+z^{2} \geq x y+y z+z x$ we get
$$
\begin{aligned}
\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} & \geq \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{2 \tau}{a b c} \\
& =\frac{\tau}{2 R E}=\frac{1}{2 R r}
\end{aligned}
$$
where we used the area formulas $E=\frac{a b c}{4 R}=\tau r$. Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{2}{r^{3} R} \geq \frac{16}{R^{4}}
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT 4. Show that there exist infinitely many positive integers $n$ such that
$$
\frac{4^{n}+2^{n}+1}{n^{2}+n+1}
$$
is an integer.
|
Solution. Let $f(n)=n^{2}+n+1$. Note that
$$
f\left(n^{2}\right)=n^{4}+n^{2}+1=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)
$$
This means that $f(n) \mid f\left(n^{2}\right)$ for every positive integer $n$. By induction on $k$, one can easily see that $f(n) \mid f\left(n^{2^{k}}\right)$ for every positive integers $n$ and $k$. Note that the required condition is equivalent to $f(n) \mid f\left(2^{n}\right)$. From the discussion above, if there exists a positive integer $n$ so that $2^{n}$ can be written as $n^{2^{k}}$, for some positive integer $k$, then $f(n) \mid f\left(2^{n}\right)$. If we choose $n=2^{2^{m}}$ and $k=2^{m}-m$ for some positive integer $m$, then $2^{n}=n^{2^{k}}$ and since there are infinitely many positive integers of the form $n=2^{2^{m}}$, we have the desired result.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A4 Real numbers $x, y, z$ satisfy
$$
0<x, y, z<1
$$
and
$$
x y z=(1-x)(1-y)(1-z) .
$$
Show that
$$
\frac{1}{4} \leq \max \{(1-x) y,(1-y) z,(1-z) x\}
$$
|
Solution: It is clear that $a(1-a) \leq \frac{1}{4}$ for any real numbers $a$ (equivalent to $0\max \{(1-x) y,(1-y) x,(1-z) x\}
$$
Now
$$
(1-x) y\frac{1}{2}$.
Using same reasoning we conclude:
$$
z\frac{1}{2}
$$
Using these facts we derive:
$$
\frac{1}{8}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}>x y z=(1-x)(1-y)(1-z)>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}
$$
Contradiction!
Remark: The exercise along with its proof generalizes for any given (finite) number of numbers, and you can consider this new form in place of the proposed one:
Exercise: If for the real numbers $x_{1}, x_{2}, \ldots, x_{n}, 0<x_{i}<1$, for all indices $i$, and
$$
x_{1} x_{2} \ldots x_{n}=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{n}\right)
$$
show that
$$
\frac{1}{4} \leq \max _{1 \leq i \leq n}\left(1-x_{i}\right) x_{i+1}
$$
(where $x_{n+1}=x_{1}$ ).
Or you can consider the following variation:
Exercise: If for the real numbers $x_{1}, x_{2}, \ldots, x_{2009}, 0<x_{i}<1$, for all indices $i$, and
$$
x_{1} x_{2} \ldots x_{2009}=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{2009}\right)
$$
show that
$$
\frac{1}{4} \leq \max _{1 \leq i \leq 2009}\left(1-x_{i}\right) x_{i+1}
$$
(where $x_{2010}=x_{1}$ ).
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A5 Let $x, y, z$ be positive real numbers. Prove that:
$$
\left(x^{2}+y+1\right)\left(x^{2}+z+1\right)\left(y^{2}+z+1\right)\left(y^{2}+x+1\right)\left(z^{2}+x+1\right)\left(z^{2}+y+1\right) \geq(x+y+z)^{6}
$$
|
Solution I: Applying Cauchy-Schwarz's inequality:
$$
\left(x^{2}+y+1\right)\left(z^{2}+y+1\right)=\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geq(x+y+z)^{2}
$$
Using the same reasoning we deduce:
$$
\left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq(x+y+z)^{2}
$$
and
$$
\left(y^{2}+x+1\right)\left(z^{2}+x+1\right) \geq(x+y+z)^{2}
$$
Multiplying these three inequalities we get the desired result.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C1 Inside of a square whose side length is 1 there are a few circles such that the sum of their circumferences is equal to 10 . Show that there exists a line that meets alt least four of these circles.
|
Solution
Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \pi$, it follows that the sum of the lengths of all mentioned projections is equal to $10 / \pi>3$. Because the side of the square is equal to 1 , we conclude that at least one point is covered with at least four of these projections. Hence, a perpendicular line to the projection side passing through this point meets at least four of the given circles, so this is a line with the desired property.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A1. Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=a b+b c+c d+d e+e a$ and $B=a c+c e+e b+b d+d a$.
Show that
$$
2005 A+B \leq 0 \text { or } \quad A+2005 B \leq 0
$$
|
## Solution
We have
$$
0=(a+b+c+d+e)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+2 A+2 B
$$
This implies that
$$
A+B \leq 0 \text { or } 2006(\dot{A}+B)=(2005 A+B)+(A+2005 B) \leq 0
$$
This implies the conclusion.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT3. Let $p$ be an odd prime. Prove that $p$ divides the integer
$$
\frac{2^{p!}-1}{2^{k}-1}
$$
for all integers $k=1,2, \ldots, p$.
|
## Solution
At first, note that $\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.
We start with the case $\mathrm{k}=\mathrm{p}$. Since $p \mid 2^{p}-2$, then $p / 22^{p}-1$ and so it suffices to prove that $p \mid 2^{(p)!}-1$. This is obvious as $p \mid 2^{p-1}-1$ and $\left(2^{p-1}-1\right) \mid 2^{(p)!}-1$.
If $\mathrm{k}=1,2, \ldots, \mathrm{p}-1$, let $m=\frac{(p-1)!}{k} \in \mathbb{N}$ and observe that $p!=k m p$. Consider $a \in \mathbb{N}$ so that $p^{a} \mid 2^{k}-1$ and observe that it suffices to prove $p^{a+1} \mid 2^{p!}-1$. The case $a=0$ is solved as the case $k=p$. If else, write $2^{k}=1+p^{a} \cdot l, l \in \mathbb{N}$ and rising at the power mp gives
$$
2^{p!}=\left(1+p^{a} \cdot l\right)^{m p}=1+m p \cdot p^{a} \cdot l+M p^{2 a}
$$
where $M n$ stands for a multiply of $\mathrm{n}$. Now it is clear that $p^{a+1} \mid 2^{p!}-1$, as claimed.
Comment. The case $\mathrm{k}=\mathrm{p}$ can be included in the case $a=0$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT5. Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$.
|
## Solution
We start with a simple fact:
Lemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.
For a proof, just note that numbers $b, 2 b, \ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is equal to 1 .
We prove that if $x, y=0,1,2, \ldots, p-1$ and $\mathrm{p}$ divides $x^{5}-y^{5}$, then $x=y$.
Indeed, assume that $x \neq y$. If $x=0$, then $p \mid y^{5}$ and so $y=0$, a contradiction.
To this point we have $x, y \neq 0$. Since
$$
p \mid(x-y)\left(x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4}\right) \text { and } p /(x-y)
$$
we have
$$
\begin{aligned}
& p l\left(x^{2}+y^{2}\right)^{2}+x y\left(x^{2}+y^{2}\right)-x^{2} y^{2} \text {, and so } \\
& p \|\left(2\left(x^{2}+y^{2}\right)+x y\right)^{2}-5 x^{2} y^{2}
\end{aligned}
$$
As $p / x y$, from the lemma we find an integer $s$ so that $s x y=k p+1, k \in \mathbb{N}$. Then
$$
p \mid\left[s\left(2 x^{2}+2 y^{2}+x y\right)\right]^{2}-5\left(k^{2} p^{2}+2 k p+1\right)
$$
and so $p \mid z^{2}-5$, where $z=s\left(2 x^{2}+2 y^{2}+x y\right)$, a contradiction.
Consequeatly $r=y$.
Since we have proved that numbers $0^{5}, 1^{5}, \ldots,(p-1)^{5}$ have distinct remainders when divided by $p$, the same goes for the numbers $0^{5}+a, 1^{5}+a, \ldots,(p-1)^{5}+a$ and the conclusion can be reached easily.
## Comments
1. For beauty we may choose $a=-2$ or any other value.
2. Moreover, we may ask only for one value of $m$, instead of "infinitely many".
3. A simple version will be to ask for a proof that the numbers $0^{5}, 1^{5}, \ldots,(p-1)^{5}$ have distinct remainders when divided by $p$.
## Combinatorics
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C1. A triangle with area 2003 is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is 2005 . Show that at mest one of the smaller triangles has area less or equal to 1.
|
## Solution
Since all the vertices are 2005 , and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least 2003. So, it follows that at least one of the small triangles has area at most 1
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A1 Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$\left(a^{5}+a^{4}+a^{3}+a^{2}+a+1\right)\left(b^{5}+b^{4}+b^{3}+b^{2}+b+1\right)\left(c^{5}+c^{4}+c^{3}+c^{2}+c+1\right) \geq 8\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
|
## Solution
We have $x^{5}+x^{4}+x^{3}+x^{2}+x+1=\left(x^{3}+1\right)\left(x^{2}+x+1\right)$ for all $x \in \mathbb{R}_{+}$.
Take $S=\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
The inequality becomes $S\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8 S$.
It remains to prove that $\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8$.
By $A M-G M$ we have $x^{3}+1 \geq 2 \sqrt{x^{3}}$ for all $x \in \mathbb{R}_{+}$.
So $\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 2^{3} \cdot \sqrt{a^{3} b^{3} c^{3}}=8$ and we are done.
Equality holds when $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A2 Let $x, y, z$ be positive real numbers. Prove that:
$$
\frac{x+2 y}{z+2 x+3 y}+\frac{y+2 z}{x+2 y+3 z}+\frac{z+2 x}{y+2 z+3 x} \leq \frac{3}{2}
$$
|
## Solution 1
Notice that $\sum_{c y c} \frac{x+2 y}{z+2 x+3 y}=\sum_{c y c}\left(1-\frac{x+y+z}{z+2 x+3 y}\right)=3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y}$.
We have to proof that $3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y} \leq \frac{3}{2}$ or $\frac{3}{2(x+y+z)} \leq \sum_{c y c} \frac{1}{z+2 x+3 y}$.
By Cauchy-Schwarz we obtain $\sum_{\text {cyc }} \frac{1}{z+2 x+3 y} \geq \frac{(1+1+1)^{2}}{\sum_{\text {cyc }}(z+2 x+3 y)}=\frac{3}{2(x+y+z)}$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A3 Let $a, b$ be positive real numbers. Prove that $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq a+b$.
|
Solution 1
Applying $x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}$ for $x=\sqrt{\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\sqrt{a b}$, we will obtain $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq \sqrt{\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \leq \sqrt{\frac{3\left(a^{2}+b^{2}+2 a b\right)}{3}}=a+b$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A6 Let $x_{i}>1$, for all $i \in\{1,2,3, \ldots, 2011\}$. Prove the inequality $\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?
|
## Solution 1
Realize that $\left(x_{i}-2\right)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4\left(x_{i}-1\right)$. So we get:
$\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. By $A M-G M$ :
$\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1} \geq 2011 \cdot \sqrt[2011]{\frac{x_{1}-1}{x_{2}-1} \cdot \frac{x_{2}-1}{x_{3}-1} \cdot \ldots \cdot \frac{x_{2011}-1}{x_{1}-1}}=2011$
Finally, we obtain that $\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 8044$.
Equality holds when $\left(x_{i}-2\right)^{2}=0,(\forall) i=\overline{1,2011}$, or $x_{1}=x_{2}=\ldots=x_{2011}=2$.
|
8044
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A7 Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$
|
## Solution 1
By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a}$.
By $A M-G M$ we have $a b+b c+c a \geq 3$ and $a+b+c \geq 3$. But $3\left(a^{2}+b^{2}+c^{2}\right) \geq(a+b+c)^{2} \geq$ $3(a+b+c)$. So $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 3+a+b+c+a b+b c+c a$. Hence $\sum_{c y c} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b+b c+c a} \geq \frac{3(a+b+c)^{2}}{(a+b+c)^{2}}=3$.
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A9 Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$.
|
## Solution 1
Case I. If $\min \left(x_{1}, x_{n}\right)=x_{1}$, we know that $x_{k} \geq \min \left(x_{k} ; x_{k+1}\right)$ for all $k \in\{1,2,3, \ldots, n-1\}$. So $x_{1}+x_{2}+\ldots+x_{n-1} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{1}$, hence $\sum_{k=2}^{n-1} x_{k} \geq 0$.
Case II. If $\min \left(x_{1}, x_{n}\right)=x_{n}$, we know that $x_{k} \geq \min \left(x_{k-1} ; x_{k}\right)$ for all $k \in\{2,3,4, \ldots, n\}$. So $x_{2}+x_{3}+\ldots+x_{n} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{n}$, hence $\sum_{k=2}^{n-1} x_{k} \geq 0$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A7 Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq(1+2 a)(1+2 b)(1+2 c)
$$
|
## Solution 1
By Cauchy-Schwarz inequality and $a b c=1$ we get
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\sqrt{\frac{1}{c a}} \cdot \sqrt{c a}\right)=(2+b c)=(2 a b c+b c)=b c(1+2 a)
\end{gathered}
$$
Analogously we get $\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right)} \geq c a(1+2 b)$ and
$\sqrt{\left(c a+a b+\frac{1}{b c}\right)\left(a b+b c+\frac{1}{c a}\right)} \geq a b(1+2 a)$.
Multiplying these three inequalities we get:
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq a^{2} b^{2} c^{2}(1+2 a)(1+2 b)(1+2 c)=
$$
$(1+2 a)(1+2 b)(1+2 c)$ because $a b c=1$.
Equality holds if and only if $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
A8 Show that
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 4\left(\frac{x}{x y+1}+\frac{y}{y z+1}+\frac{z}{z x+1}\right)^{2}
$$
for any real positive numbers $x, y$ and $z$.
|
## Solution
The idea is to split the inequality in two, showing that
$$
\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
can be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
On the other hand, as
$$
\sqrt{\frac{x}{y}} \geq \frac{2 x}{x y+1} \Leftrightarrow(\sqrt{x y}-1)^{2} \geq 0
$$
by summation one has
$$
\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}} \geq \frac{2 x}{x y+1}+\frac{2 y}{y z+1}+\frac{2 z}{z x+1}
$$
The rest is obvious.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C1 On a $5 \times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an empty square such that none of the neighboring squares contains a white marker (two squares are called neighboring if they contain a common side). If it is possible for the child to succeed in coloring all the markers black, we say that the initial positioning of the markers was good.
a) Prove that if $n=20$, then a good initial positioning exists.
b) Prove that if $n=21$, then a good initial positioning does not exist.
|
Solution
a) Position 20 white markers on the board such that the left-most column is empty. This
positioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on position $(i, j-1)$.
b) Suppose there exists a good positioning with 21 white markers on the board i.e. there exists a re-coloring of them all, one by one. In any moment when there are 21 markers on the board, there must be at least one column completely filled with markers, and there must be at least one row completely filled with markers. So, there exists a "cross" of markers on the board. At the initial position, each such cross is completely white, at the final position each such cross is completely black, and at every moment when there are 21 markers on the board, each such cross is monochromatic. But this cannot be, since every two crosses have at least two common squares and therefore it is not possible for a white cross to vanish and for a black cross to appear by re-coloring of only one marker. Contradiction!
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false |
C3 Integers $1,2, \ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.
|
## Solution
Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.
For our assumption,
$$
S \equiv 0+1+\ldots+2 n-1=\frac{(2 n-1) 2 n}{2}=(2 n-1) n \equiv n \quad(\bmod 2 n)
$$
But, if we sum, breaking all sums into its components, we derive
$$
S \equiv 2(1+\ldots+2 n)=2 \cdot \frac{2 n(2 n+1)}{2}=2 n(2 n+1) \equiv 0 \quad(\bmod 2 n)
$$
From the last two conclusions we derive $n \equiv 0(\bmod 2 n)$. Contradiction.
Therefore, there are two sums with the same remainder modulo $2 n$.
Remark: The result is no longer true if one replaces $2 n$ by $2 n+1$. Indeed, one could assign the number $k$ to the box labeled $k$, thus obtaining the sums $2 k, k=\overline{1,2 n+1}$. Two such numbers give different remainders when divided by $2 n+1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
G2 For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$ ), a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.
|
## Solution
Consider the points $M^{\prime}$ on the ray $B A$ (after $A$ ), $N^{\prime}$ on the ray $C B$ (after $B$ ) and $P^{\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\prime}, B N=B N^{\prime}, C P=C P^{\prime}$. Since $A M-B C=B N-A C=B N^{\prime}-A C$, we get $C M=A C+A M=B C+B N^{\prime}=C N^{\prime}$. Thus triangle $M C N^{\prime}$ is isosceles, so the perpendicular bisector of $\left[M N^{\prime}\right]$ bisects angle $A C B$ and hence passes through the incenter $I$ of triangle $A B C$. Arguing similarly, we may conclude that $I$ lies also on the perpendicular bisectors of $\left[N P^{\prime}\right]$ and $\left[P M^{\prime}\right]$. On the other side, $I$ clearly lies on the perpendicular bisectors of $\left[M M^{\prime}\right],\left[N N^{\prime}\right]$ and $\left[P P^{\prime}\right]$. Thus the hexagon $M^{\prime} M N^{\prime} N P^{\prime} P$ is cyclic. Then angle $P M N$ equals angle $P N^{\prime} N$, which measures $90^{\circ}-\frac{\beta}{2}$ (the angles of triangle $A B C$ are $\alpha, \beta, \gamma$ ). In the same way angle $M N P$ measures $90^{\circ}-\frac{\gamma}{2}$ and angle $M P N$ measures $90^{\circ}-\frac{\alpha}{2}$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT2 Let $n \geq 2$ be a fixed positive integer. An integer will be called " $n$-free" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.
|
## Solution
We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \ldots, \frac{p_{k}}{q_{k}}, \ldots$, with $\left(p_{k}, q_{k}\right)=1$ and $q_{k}>1$ for each $k$. Let $\frac{p}{q}=\frac{p_{1} p_{2} \ldots p_{n-1}}{q_{1} q_{2} \ldots q_{n-1}}$, where $(p, q)=1$. For each $i \geq n$, the number $\frac{p}{q} \cdot \frac{p_{i}}{q_{i}}$ is an integer, so $q_{i}$ is a divisor of $p$ (as $q_{i}$ and $p_{i}$ are coprime). But $p$ has a finite set of divisors, so there are $n$ numbers of $M$ with equal denominators. Their product cannot be an integer, a contradiction.
Now suppose that $M$ contains a fraction $\frac{a}{b}$ in lowest terms with $b>1$. Take a prime divisor $p$ of $b$. If we take any $n-1$ integers from $M$, their product with $\frac{a}{b}$ is an integer, so some of them is a multiple of $p$. Therefore there are infinitely many multiples of $p$ in $M$, and the product of $n$ of them is not $n$-free, a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
NT10 Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.
|
## Solution
If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.
Perfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).
If $n=3 k$ then $2^{3 k}+3^{2 k}>\left(3^{k}\right)^{3}$. Also, $\left(3^{k}+1\right)^{3}=3^{3 k}+3 \cdot 3^{2 k}+3 \cdot 3^{k}+1>3^{3 k}+3^{2 k}=$ $3^{3 k}+9^{k}>3^{3 k}+8^{k}=3^{3 k}+2^{3 k}$. But, $3^{k}$ and $3^{k}+1$ are two consecutive integers so $2^{3 k}+3^{3 k}$ is not a perfect cube.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 1. A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.
|
## Solution.
$$
\begin{aligned}
A & =\underbrace{44 \ldots 44}_{2 n}=\underbrace{44 \ldots 4}_{n} \underbrace{44 \ldots 4}_{n}=\underbrace{44 \ldots 4}_{n} \underbrace{400 \ldots 0}_{n}-\underbrace{44 \ldots 4}_{n}+\underbrace{88 \ldots 8}_{n}=\underbrace{44 \ldots 4}_{n} \cdot\left(10^{n}-1\right)+B \\
& =4 \cdot \underbrace{11 \ldots 1}_{n} \cdot \underbrace{99 \ldots 9}_{n}+B=2^{2} \cdot \underbrace{11 \ldots 1}_{n} \cdot 3^{2} \cdot \underbrace{11 \ldots 1}_{n}+B=\underbrace{66}_{n} \ldots 6 \\
& =[\frac{36}{4} \cdot \underbrace{88 \ldots 8}_{n}+B=[3 \cdot \underbrace{22 \ldots 2}_{n}]^{2}+B=\left(\frac{3}{4} B\right)^{2}+B .
\end{aligned}
$$
So,
$$
\begin{aligned}
A+2 B+4 & =\left(\frac{3}{4} B\right)^{2}+B+2 B+4=\left(\frac{3}{4} B\right)^{2}+2 \cdot \frac{3}{4} B \cdot 2+2^{2}=\left(\frac{3}{4} B+2\right)^{2}=(\frac{3}{4} \cdot \underbrace{88 \ldots 8}_{n}+2)^{2} \\
& =(3 \cdot \underbrace{22 \ldots 2}_{n}+2)^{2}=\underbrace{66 \ldots 68^{2}}_{n-1}
\end{aligned}
$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 2. Let $a, b, c$ be lengths of triangle sides, $p=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $q=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$.
Prove that $|p-q|<1$.
|
Solution: One has
$$
\begin{aligned}
a b c|p-q| & =a b c\left|\frac{c-b}{a}+\frac{a-c}{b}+\frac{b-a}{c}\right| \\
& =\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\right|= \\
& =\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\right|= \\
& =\left|(b-c)\left(a c-a^{2}-b c+a b\right)\right|= \\
& =|(b-c)(c-a)(a-b)| .
\end{aligned}
$$
Since $|b-c|<a,|c-a|<b$ and $|a-b|<c$ we infere
$$
|(b-c)(c-a)(a-b)|<a b c
$$
and
$$
|p-q|=\frac{|(b-c)(c-a)(a-b)|}{a b c}<1
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 3: Let $a, b, c$ be real numbers such that $a^{2}+b^{2}+c^{2}=1$. Prove that. $P=a b+b c+c a-2(a+b+c) \geq-\frac{5}{2}$. Are there values of $a, b, c$, such that $P=-\frac{5}{2}$.
|
Solution: We have $a b+b c+c a=\frac{(a+b+c)^{2}-c^{2}-b^{2}-a^{2}}{2}=\frac{(a+b+c)^{2}-1}{2}$.
If put $t=a+b+c$ we obtain
$$
P=\frac{t^{2}-1}{2}-2 t=\frac{t^{2}-4 t-1}{2}=\frac{(t-2)^{2}-5}{2} \geq-\frac{5}{2}
$$
Obviously $P=-\frac{5}{2}$ when $t=2$, i.e. $a+b+c=2$, or $c=2-a-b$. Substitute in $a^{2}+b^{2}+c^{2}=1$ and obtain $2 a^{2}+2(b-2) a+2 b^{2}-4 b+3=0$. Sinse this quadratic equation has solutions it follows that $(b-2)^{2}-2\left(2 b^{2}-3 b+3\right) \geq 0$, from where
$$
-3 b^{2}+4 b-6 \geq 0
$$
or
$$
3 b^{2}-4 b+6 \leq 0
$$
But $3 b^{2}-4 b+6=3\left(b-\frac{2}{3}\right)^{2}+\frac{14}{3}>0$. The contradiction shows that $P=-\frac{5}{2}$.
Comment: By the Cauchy Schwarz inequality $|t| \leq \sqrt{3}$, so the smallest value of $P$ is attained at $t=\sqrt{3}$ and equals $1-2 \sqrt{3} \approx-2.46$.
|
P\geq-\frac{5}{2}
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## ALG 4.
Let $a, b, c$ be rational numbers such that
$$
\frac{1}{a+b c}+\frac{1}{b+a c}=\frac{1}{a+b}
$$
Prove that $\sqrt{\frac{c-3}{c+1}}$ is also a rational number
|
Solution. By cancelling the denominators
$$
(a+b)^{2}(1+c)=a b+c\left(a^{2}+b^{2}\right)+a b c^{2}
$$
and
$$
a b(c-1)^{2}=(a+b)^{2}
$$
If $c=-1$, we obtrin the contradiction
$$
\frac{1}{a-b}+\frac{1}{b-a}=\frac{1}{a+b}
$$
Furtherrdore,
$$
\begin{aligned}
(c-3)(c+1) & =(c-1)^{2}-4=\frac{(a+b)^{2}}{a b}-4 \\
& =\frac{(a-b)^{2}}{a b}=\left(\frac{(a-b)(c-1)}{a+b}\right)^{2}
\end{aligned}
$$
Thus
$$
\sqrt{\frac{c-3}{c+1}}=\frac{\sqrt{(c-3)(c+1)}}{c+1}=\frac{|a-b||c-1|}{(c+1)|a+b|} \in \mathrm{Q}
$$
as needed.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 5. Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that
$$
\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right| \geq 2
$$
|
Solution. Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have
$$
\begin{aligned}
E= & \left(a b c-c^{2} a\right)+\left(c a^{2}-a^{2} b\right)+\left(b c^{2}-b^{2} c\right)+\left(a b^{2}-a b c\right)= \\
& (b-c)\left(a c-a^{2}-b c+a b\right)=(b-c)\left(a a^{2}-b\right)(c-a)
\end{aligned}
$$
So, $|E|=|a-b| \cdot|b-c| \cdot|c-a|$. By hypothesis each factor from $|E|$ is a positive integer. We shall prove that at least one factor from $|E|$ is greater than 1. Suppose that $|a-b|=|b-c|=|c-a|=1$. It follows that the numbers $a-b, b-c, c-a$ are odd. So, the number $0=(a-b)+(b-c) \div(c-a)$ is olso odd, a contradiction. Hence, $|E| \geq 1 \cdot 1 \cdot 2=2$.
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 6'. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
a+b+c \geq a b c+2
$$
|
Solution. Eliminating $c$ gives
$$
a+b+c-a b c=a+b+(1-a b) c=a+b+\frac{(1-a b)(3-a b)}{a+b}
$$
Put $x=\sqrt{a b}$. Then $a+b \geq 2 x$, and since $1<x^{2}<3, \frac{(1-a b)(3-a b)}{a+b} \geq \frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}$.
It then suffices to prove that
$$
2 x+\frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x} \geq 2
$$
This iast inequality follows from the arithrnelic-geomeric means inequadily
$$
2 x+\frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}=\frac{3+x^{4}}{2 x}=\frac{1}{2 x}+\frac{1}{2 x}+\frac{1}{2 x}+\frac{x^{3}}{2} \geq 4\left(\frac{1}{-16}\right)^{\frac{1}{4}}=2
$$
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 7 .
Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
|
Solution. We have $y \leq \frac{1+y^{2}}{2}$, hence $\quad$
$$
\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+x^{2}}{1+z^{2}+\frac{1+\dot{y}^{2}}{2}}
$$
and the similar inequalities.
Setting $a=1+x^{2}, b=1+y^{2}, c=1+z^{2}$, it sufices to prove that
$$
\frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \geq 1
$$
for all $a, b, c \geq 0$.
Put $A=2 c+b, B=2 a+c, C=2 b+a$. Then
$$
a=\frac{C+4 B-2 A}{9}, b=\frac{A+4 C-2 B}{9}, c=\frac{B+4 A-2 C}{9}
$$
and (1) rewrites as
$$
\frac{C+4 B-2 A}{A}+\frac{A+4 C-2 B}{B}+\frac{B+4 A-2 C}{C} \geq 9
$$
and consequently
$$
\frac{C}{A}+\frac{A}{B}+\frac{B}{C}+4\left(\frac{B}{A}+\frac{C}{B}+\frac{A}{C}\right) \geq 15
$$
As $A, B, C>0$, by $A M-G M$ inequality we have
$$
\frac{A}{B}+\frac{B}{C}+\frac{C}{A} \geq 3 \sqrt[3]{\frac{A}{B} \cdot \frac{B}{C} \cdot \frac{C}{A}}
$$
and
$$
\frac{B}{A}+\frac{C}{B}+\frac{A}{C} \geq 3
$$
and we are done.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
ALG 8. Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.
|
Solution. Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c
$$
a triangle with such length sides there exist. After the simple calculations we have
$$
\begin{gathered}
A=\left\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\right\} \\
B=\left\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\right\}
\end{gathered}
$$
It easy to prove that
$$
\begin{gathered}
x+y+z=m+n+p=3\left[(k+1)^{2}+(k+2)^{2}+(k+3)^{2}\right] \\
x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}=9\left[(k+1)^{4}+(k+2)^{4}+(k+3)^{4}\right]
\end{gathered}
$$
$>$ From the inequality $3(k+1)^{2}-2>2003$ we obtain $k \geq 25$. For $k=25$ we have an example of two sets
$$
A=\{2026,2191,2350\}, \quad B=\{2028,2187,2352\}
$$
with desired properties.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
COM 3. Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .
|
Solution: Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see that amongst any 9 natural numbers there are 5 such that sum of them is divisible by 5 . Since we have 9 numbers, all of them are divisible by 3 , there are 5 such that sum of them is divisible by 15 .
## $\operatorname{COM} 4$.
$n$ points are given in a plane, not three of them colinear. One observes that no matter how we label the points from 1 to $n$, the broken line joining the points $1,2,3, \ldots, n$ (in this order) do not intersect itself.
Find the maximal value of $n$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false |
COM 5. If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.
|
Solution. Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.
First assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose that the segment $X Y$ with length 2 have the midpoint $T$ such that the points $X, Y, T$ have the same colour (for example, red). We construct the equilateral triangle. $X Y Z$. Hence, the point $Z$ is blue. Let $U$ and $V$ be the midpoints of the segments $X Z$ and $Y Z$ respectively. So, the points $U$ and $V$ are blue. We obtain a contradiction, because the equilateral triangle $U V Z$ have three blue vertices.
Second assertion: in the same way we prove that in the plane there no exists a segment with the length 4 such that the ends and the midpoint of this segment have the same colour.
Consider the equilateral triangle $A B C$ with length side 4 and divide it into 16 equilateral triangles with length sides 1. L $0: D$ be the midpoint of the segment $A B$. The vertices $A, B, C$ don't have the same colour. WLOG we suppose that $A$ and $B$ are red and $C$ is blue. So, the point $D$ is blue too. We shall investigate the following cases:
a) The midpoints $E$ and $F$ of the sides $A C$ and, respectively, $B C$ are red. From the first assertion it follows that the midpoints $M$ and $N$ of the segments $A E$ and, respectively, $B F$ are blue. Hence, the equilateral triangle $M N C$ have three blue vertices, a contradiction.
b) Let $E$ is red and $F$ is blue. The second one position of $E$ and $F$ is simmetrical. If $P, K, L$ are the midpoints of the segments $C F, A D, B D$ respectively, then by first assertion $P$ is red, $M$ is blue and $N$ is red. This imply that $K$ and $L$ are blue. So, the segment $K L$ with length 2 has the blue ends and blue midpoint, a contradiction.
c) If $E$ and $F$ are blue, then the equilateral triangle $E F C$ has three blue vertices, a contradiction.
Hence, in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m \in\{1,2,3,4\}$.
Comment: The formuation of the problem suggests that one has to find 4 triangles, one for each $m$ from the set $\{1,2,3,4\}$ whereas the solution is for one $m$. A better formulation is:
Each point of the plane is painted in red or blue. Prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m$ is some number from the set $\{1,2,3,4\}$.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false |
## GEO 3.
Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\prime}$. Prove that $G, B, C, A^{\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.
|
Solution. Observe first that $G A \perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,
$$
G A \perp G C \Leftrightarrow \frac{4}{9} m_{a}^{2}+\frac{4}{9} m_{c}^{2}=b^{2} \Leftrightarrow 5 b^{2}=a^{2}+c^{2}
$$
Moreover,
$$
G B^{2}=\frac{4}{9} m_{b}^{2}=\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\frac{9 b^{2}}{9}=b^{2}
$$
hence $G B=A C=C A^{\prime}$ (1). Let $C^{\prime}$ be the intersection point of the lines $G C$ and $A B$. Then $C C^{\prime}$ is the middle line of the triangle $A B A^{\prime}$, hence $G C \| B A^{\prime}$. Consequently, $G C A^{\prime} B$ is a trapezoid. From (1) we find that $G C A^{\prime} B$ is isosceles, thus cyclic, as needed.
Conversely, since $G C A^{\prime} B$ is a cyclic trapezoid, then it is also isosceles. Thus $C A^{\prime}=$ $G B$, which leads to (1).
Comment: An alternate proof is as follows:
Let $M$ be the midpoint of $A C$. Then the triangles $M C G$ and $M A^{\prime} B$ are similar. So $G C$ is parallel to $A^{\prime} B$.
$G A \perp G C$ if and only if $G M=M C$. By the above similarity, this happen if and only if $A^{\prime} C=G B$; if and only if the trapezoid is cyclic.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
87.2. Let $A B C D$ be a parallelogram in the plane. We draw two circles of radius $R$, one through the points $A$ and $B$, the other through $B$ and $C$. Let $E$ be the other point of
intersection of the circles. We assume that $E$ is not a vertex of the parallelogram. Show that the circle passing through $A, D$, and $E$ also has radius $R$.
|
Solution. (See Figure 1.) Let $F$ and $G$ be the centers of the two circles of radius $R$ passing through $A$ and $B$; and $B$ and $C$, respectively. Let $O$ be the point for which the the rectangle $A B G O$ is a parallelogram. Then $\angle O A D=\angle G B C$, and the triangles $O A D$ and $G B C$ are congruent (sas). Since $G B=G C=R$, we have $O A=O D=R$. The quadrangle $E F B G$ is a rhombus. So $E F\|G B\| O A$. Moreover, $E F=O A=R$, which means that $A F E O$ is a parallelogram. But this implies $O E=A F=R$. So $A, D$, and $E$ all are on the circle of radius $R$ centered at $O$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false |
87.4. Let $a, b$, and $c$ be positive real numbers. Prove:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
|
Solution. The arithmetic-geometric inequality yields
$$
3=3 \sqrt[3]{\frac{a^{2}}{b^{2}} \cdot \frac{b^{2}}{c^{2}} \cdot \frac{c^{2}}{a^{2}}} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
or
$$
\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
$$
On the other hand, the Cauchy-Schwarz inequality implies
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+ & \frac{c}{a} \leq \sqrt{1^{2}+1^{2}+1^{2}} \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}} \\
& =\sqrt{3} \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
\end{aligned}
$$
We arrive at the inequality of the problem by combining (1) and (2).
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
88.2. Let $a, b$, and $c$ be non-zero real numbers and let $a \geq b \geq c$. Prove the inequality
$$
\frac{a^{3}-c^{3}}{3} \geq a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
When does equality hold?
|
Solution. Since $c-b \leq 0 \leq a-b$, we have $(a-b)^{3} \geq(c-b)^{3}$, or
$$
a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \geq c^{3}-3 b c^{2}+3 b^{2} c-b^{3}
$$
On simplifying this, we immediately have
$$
\frac{1}{3}\left(a^{3}-c^{3}\right) \geq a^{2} b-a b^{2}+b^{2} c-b c^{2}=a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
A sufficient condition for equality is $a=c$. If $a>c$, then $(a-b)^{3}>(c-b)^{3}$, which makes the proved inequality a strict one. So $a=c$ is a necessary condition for equality, too.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false |
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