Found problems: 112
2021 IMO, 2
Show that the inequality \[\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|}\leqslant \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|}\]holds for all real numbers $x_1,\ldots x_n.$
2020 IMO Shortlist, A7
Let $n$ and $k$ be positive integers. Prove that for $a_1, \dots, a_n \in [1,2^k]$ one has
\[ \sum_{i = 1}^n \frac{a_i}{\sqrt{a_1^2 + \dots + a_i^2}} \le 4 \sqrt{kn}. \]
2014 Iran Team Selection Test, 5
$n$ is a natural number. for every positive real numbers $x_{1},x_{2},...,x_{n+1}$ such that $x_{1}x_{2}...x_{n+1}=1$ prove that:
$\sqrt[x_{1}]{n}+...+\sqrt[x_{n+1}]{n} \geq n^{\sqrt[n]{x_{1}}}+...+n^{\sqrt[n]{x_{n+1}}}$
2016 Croatia Team Selection Test, Problem 1
Let $n \ge 1$ and $x_1, \ldots, x_n \ge 0$. Prove that
$$ (x_1 + \frac{x_2}{2} + \ldots + \frac{x_n}{n}) (x_1 + 2x_2 + \ldots + nx_n) \le \frac{(n+1)^2}{4n} (x_1 + x_2 + \ldots + x_n)^2 .$$
2020 China Second Round Olympiad, 2
Let $n\geq3$ be a given integer, and let $a_1,a_2,\cdots,a_{2n},b_1,b_2,\cdots,b_{2n}$ be $4n$ nonnegative reals, such that $$a_1+a_2+\cdots+a_{2n}=b_1+b_2+\cdots+b_{2n}>0,$$ and for any $i=1,2,\cdots,2n,$ $a_ia_{i+2}\geq b_i+b_{i+1},$ where $a_{2n+1}=a_1,$ $a_{2n+2}=a_2,$ $b_{2n+1}=b_1.$ Detemine the minimum of $a_1+a_2+\cdots+a_{2n}.$
1971 IMO Longlists, 21
Let \[ E_n=(a_1-a_2)(a_1-a_3)\ldots(a_1-a_n)+(a_2-a_1)(a_2-a_3)\ldots(a_2-a_n)+\ldots+(a_n-a_1)(a_n-a_2)\ldots(a_n-a_{n-1}). \] Let $S_n$ be the proposition that $E_n\ge0$ for all real $a_i$. Prove that $S_n$ is true for $n=3$ and $5$, but for no other $n>2$.
2004 All-Russian Olympiad, 4
Let $n > 3$ be a natural number, and let $x_1$, $x_2$, ..., $x_n$ be $n$ positive real numbers whose product is $1$. Prove the inequality \[ \frac {1}{1 + x_1 + x_1\cdot x_2} + \frac {1}{1 + x_2 + x_2\cdot x_3} + ... + \frac {1}{1 + x_n + x_n\cdot x_1} > 1. \]
2004 IMO Shortlist, 1
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
2023 Bulgaria National Olympiad, 5
For every positive integer $n$ determine the least possible value of the expression
\[|x_{1}|+|x_{1}-x_{2}|+|x_{1}+x_{2}-x_{3}|+\dots +|x_{1}+x_{2}+\dots +x_{n-1}-x_{n}|\]
given that $x_{1}, x_{2}, \dots , x_{n}$ are real numbers satisfying $|x_{1}|+|x_{2}|+\dots+|x_{n}| = 1$.
1997 Estonia Team Selection Test, 2
Prove that for all positive real numbers $a_1,a_2,\cdots a_n$ \[\frac{1}{\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots +\frac{1}{1+a_n}}-\frac{1}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_n}}\geq \frac{1}{n}\] When does the inequality hold?
2011 Korea Junior Math Olympiad, 7
For those real numbers $x_1 , x_2 , \ldots , x_{2011}$ where each of which satisfies $0 \le x_1 \le 1$ ($i = 1 , 2 , \ldots , 2011$), find the maximum of
\[ x_1^3+x_2^3+ \cdots + x_{2011}^3 - \left( x_1x_2x_3 + x_2x_3x_4 + \cdots + x_{2011}x_1x_2 \right) \]
1966 IMO Longlists, 26
Prove the inequality
[b]a.)[/b] $
\left( a_{1}+a_{2}+...+a_{k}\right) ^{2}\leq k\left(
a_{1}^{2}+a_{2}^{2}+...+a_{k}^{2}\right) , $
where $k\geq 1$ is a natural number and $a_{1},$ $a_{2},$ $...,$ $a_{k}$ are arbitrary real numbers.
[b]b.)[/b] Using the inequality (1), show that if the real numbers $a_{1},$ $a_{2},$ $...,$ $a_{n}$ satisfy the inequality
\[
a_{1}+a_{2}+...+a_{n}\geq \sqrt{\left( n-1\right) \left(
a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\right) },
\]
then all of these numbers $a_{1},$ $a_{2},$ $\ldots,$ $a_{n}$ are non-negative.
1998 IMO Shortlist, 1
Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that
\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]
2008 Mathcenter Contest, 3
Let ${a_1,a_2,\dots,a_n}$ be positive real numbers, ${n>1}$. Denote by $g_n$ their geometric mean, and by $A_1,A_2,\dots,A_n$ the sequence of arithmetic means defined by \[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \] Let $G_n$ be the geometric mean of $A_1,A_2,\dots,A_n$. Prove the inequality \[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \] and establish the cases of equality.
[i]Proposed by Finbarr Holland, Ireland[/i]
1996 Taiwan National Olympiad, 2
Let $0<a\leq 1$ be a real number and let $a\leq a_{i}\leq\frac{1}{a_{i}}\forall i=\overline{1,1996}$ are real numbers. Prove that for any nonnegative real numbers $k_{i}(i=1,2,...,1996)$ such that $\sum_{i=1}^{1996}k_{i}=1$ we have $(\sum_{i=1}^{1996}k_{i}a_{i})(\sum_{i=1}^{1996}\frac{k_{i}}{a_{i}})\leq (a+\frac{1}{a})^{2}$.
2001 Kazakhstan National Olympiad, 3
For positive numbers $ x_1, x_2, \ldots, x_n $ $ (n \geq 1) $ the following equality holds $$ \frac {1} {{1 + x_1}} + \frac {1} {{1 + x_2}} + \ldots + \frac {1} {{1 + x_n}} = 1. $$ Prove that $ x_1 \cdot x_2 \cdot \ldots \cdot x_n \geq (n-1) ^ n. $
2004 IMO Shortlist, 7
Let ${a_1,a_2,\dots,a_n}$ be positive real numbers, ${n>1}$. Denote by $g_n$ their geometric mean, and by $A_1,A_2,\dots,A_n$ the sequence of arithmetic means defined by \[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \] Let $G_n$ be the geometric mean of $A_1,A_2,\dots,A_n$. Prove the inequality \[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \] and establish the cases of equality.
[i]Proposed by Finbarr Holland, Ireland[/i]
2011 Belarus Team Selection Test, 3
Let $x_1, \ldots , x_{100}$ be nonnegative real numbers such that $x_i + x_{i+1} + x_{i+2} \leq 1$ for all $i = 1, \ldots , 100$ (we put $x_{101 } = x_1, x_{102} = x_2).$ Find the maximal possible value of the sum $S = \sum^{100}_{i=1} x_i x_{i+2}.$
[i]Proposed by Sergei Berlov, Ilya Bogdanov, Russia[/i]
2005 Romania National Olympiad, 4
Let $x_1,x_2,\ldots,x_n$ be positive reals. Prove that
\[ \frac 1{1+x_1} + \frac 1{1+x_1+x_2} + \cdots + \frac 1{1+x_1+\cdots + x_n} < \sqrt { \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}} . \]
[i]Bogdan Enescu[/i]
2005 Germany Team Selection Test, 2
Let $ n$ be a positive integer such that $ n\geq 3$. Let $ a_1$, $ a_2$, ..., $ a_n$ and $ b_1$, $ b_2$, ..., $ b_n$ be $ 2n$ positive real numbers satisfying the equations
\[ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 1, \quad \text{and} \quad b_1^2 \plus{} b_2^2 \plus{} ... \plus{} b_n^2 \equal{} 1.\]
Prove the inequality
\[a_1\left(b_1 \plus{} a_2\right) \plus{} a_2\left(b_2 \plus{} a_3\right) \plus{} ... \plus{} a_{n \minus{} 1}\left(b_{n \minus{} 1} \plus{} a_n\right) \plus{} a_n\left(b_n \plus{} a_1\right) < 1.\]
1995 IMO Shortlist, 3
Let $ n$ be an integer, $ n \geq 3.$ Let $ a_1, a_2, \ldots, a_n$ be real numbers such that $ 2 \leq a_i \leq 3$ for $ i \equal{} 1, 2, \ldots, n.$ If $ s \equal{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n,$ prove that
\[ \frac{a^2_1 \plus{} a^2_2 \minus{} a^2_3}{a_1 \plus{} a_2 \minus{} a_3} \plus{} \frac{a^2_2 \plus{} a^2_3 \minus{} a^2_4}{a_2 \plus{} a_3 \minus{} a_4} \plus{} \ldots \plus{} \frac{a^2_n \plus{} a^2_1 \minus{} a^2_2}{a_n \plus{} a_1 \minus{} a_2} \leq 2s \minus{} 2n.\]
2010 Danube Mathematical Olympiad, 5
Let $n\ge3$ be a positive integer. Find the real numbers $x_1\ge0,\ldots,x_n\ge 0$, with $x_1+x_2+\ldots +x_n=n$, for which the expression \[(n-1)(x_1^2+x_2^2+\ldots+x_n^2)+nx_1x_2\ldots x_n\] takes a minimal value.
2005 Korea Junior Math Olympiad, 7
If positive reals $ x_1,x_2,\cdots,x_n $ satisfy $\sum_{i=1}^{n}x_i=1.$ Prove that$$\sum_{i=1}^{n}\frac{1}{1+\sum_{j=1}^{i}x_j}<\sqrt{\frac{2}{3}\sum_{i=1}^{n}\frac{1}{x_i}}
$$
2012 IMO Shortlist, A3
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]
[i]Proposed by Angelo Di Pasquale, Australia[/i]
2014 Taiwan TST Round 2, 1
Let $a_i > 0$ for $i=1,2,\dots,n$ and suppose $a_1 + a_2 + \dots + a_n = 1$. Prove that for any positive integer $k$,
\[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \]