This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 112

2005 Austrian-Polish Competition, 3
Tags: n-variable inequality , inequalities
Let $a_0, a_1, a_2, ... , a_n$ be real numbers, which fulfill the following two conditions: a) $0 = a_0 \leq a_1 \leq a_2 \leq ... \leq a_n$. b) For all $0 \leq i < j \leq n$ holds: $a_j - a_i \leq j-i$. Prove that $$\left( \displaystyle \sum_{i=0}^n a_i \right)^2 \geq \sum_{i=0}^n a_i^3.$$

2011 Morocco TST, 2
Tags: inequalities , n-variable inequality
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]

2014 Taiwan TST Round 2, 1
Tags: function , calculus , inequalities , derivative , n-variable inequality
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 $k$, \[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \]

2004 China National Olympiad, 2
Tags: polynomial , inequalities , algebra , n-variable inequality
For a given positive integer $n\ge 2$, suppose positive integers $a_i$ where $1\le i\le n$ satisfy $a_1<a_2<\ldots <a_n$ and $\sum_{i=1}^n \frac{1}{a_i}\le 1$. Prove that, for any real number $x$, the following inequality holds \[\left(\sum_{i=1}^n\frac{1}{a_i^2+x^2}\right)^2\le\frac{1}{2}\cdot\frac{1}{a_1(a_1-1)+x^2} \] [i]Li Shenghong[/i]

1995 Poland - Second Round, 4
Tags: algebra , inequalities , n-variable inequality , sum
Positive real numbers $x_1,x_2,...,x_n$ satisfy the condition $\sum_{i=1}^n x_i \le \sum_{i=1}^n x_i ^2$ . Prove the inequality $\sum_{i=1}^n x_i^t \le \sum_{i=1}^n x_i ^{t+1}$ for all real numbers $t > 1$.

2023 Bulgaria National Olympiad, 5
Tags: algebra , inequalities , n-variable inequality
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 $|x_{1}|+|x_{2}|+\dots+|x_{n}| = 1$.

2014 Singapore MO Open, 3
Tags: inequalities , n-variable inequality
Let $0<a_1<a_2<\cdots <a_n$ be real numbers. Prove that \[\left (\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots +\frac{1}{1+a_n}\right )^2 \leq \frac{1}{a_1}+\frac{1}{a_2-a_1}+\cdots +\frac{1}{a_n-a_{n-1}}.\]

1998 Singapore Team Selection Test, 2
Tags: inequalities , algebra , n-variable inequality
Let $ a_1\geq \cdots \geq a_n \geq a_{n \plus{} 1} \equal{} 0$ be real numbers. Show that \[ \sqrt {\sum_{k \equal{} 1}^n a_k} \leq \sum_{k \equal{} 1}^n \sqrt k (\sqrt {a_k} \minus{} \sqrt {a_{k \plus{} 1}}). \] [i]Proposed by Romania[/i]

2010 IMO Shortlist, 3
Tags: inequalities , n-variable inequality
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]

2011 ISI B.Stat Entrance Exam, 1
Tags: calculus , inequalities , rearrangement inequality , n-variable inequality
Let $x_1, x_2, \cdots , x_n$ be positive reals with $x_1+x_2+\cdots+x_n=1$. Then show that \[\sum_{i=1}^n \frac{x_i}{2-x_i} \ge \frac{n}{2n-1}\]

2006 Lithuania Team Selection Test, 1
Tags: inequalities , n-variable inequality
Let $a_1, a_2, \dots, a_n$ be positive real numbers, whose sum is $1$. Prove that \[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{1}{2} \]

2014 Iran Team Selection Test, 5
Tags: function , inequalities , n-variable inequality
$n$ 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: $\sqrt[x_{1}]{n}+...+\sqrt[x_{n+1}]{n} \geq n^{\sqrt[n]{x_{1}}}+...+n^{\sqrt[n]{x_{n+1}}}$