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

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Found problems: 112

Let $n \ge 1$ and $x_1, \ldots, x_n \ge 0$. Prove that $$ (x_1 + \frac{x_2}{2} + \ldots + \frac{x_n}{n}) (x_1 + 2x_2 + \ldots + nx_n) \le \frac{(n+1)^2}{4n} (x_1 + x_2 + \ldots + x_n)^2 .$$
Let $n \geqslant 2$ be an integer and let $1<a_1 \le a_2 \le \dots \le a_n$ be $n$ real numbers such that $a_1+a_2+\dots+a_n=2n$. Prove that$$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.$$ [i]Proposed by Kapil Pause[/i]
Let $x_1, x_2, \cdots , x_n$ be positive numbers. Prove that \[\frac{x_1^2}{x_1^2+x_2x_3} + \frac{x_2^2}{x_2^2+x_3x_4} + \cdots +\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1} +\frac{x_n^2}{x_n^2+x_1x_2} \leq n-1\]
Let $n\ge2$ be a natural number, and $a_i$ be positive numbers, where $i=1,2,\cdots,n.$ Show that \[\left(a_1^3+1\right)\left(a_2^3+1\right)\cdots\left(a_n^3+1\right) \geq \left(a_1^2a_2+1\right)\left(a_2^2a_3+1\right)\cdots\left(a_n^2a_1+1\right)\]
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 $n>2$.
For real numbers $ x_1, x_2, \dots, x_n $ and $ y_1, y_2, \dots, y_n $ , the inequalities hold $ x_1 \geq x_2 \geq \ldots \geq x_n> 0 $ and $$ y_1 \geq x_1, ~ y_1y_2 \geq x_1x_2, ~ \dots, ~ y_1y_2 \dots y_n \geq x_1x_2 \dots x_n. $$ Prove that $ ny_1 + (n-1) y_2 + \dots + y_n \geq x_1 + 2x_2 + \dots + nx_n $.
Let $x_1 \le x_2 \le \ldots \le x_{2n-1}$ be real numbers whose arithmetic mean equals $A$. Prove that $$2\sum_{i=1}^{2n-1}\left( x_{i}-A\right)^2 \ge \sum_{i=1}^{2n-1}\left( x_{i}-x_{n}\right)^2.$$
Given positive real numbers $a_1, a_2, \dots, a_n$. Let \[ m = \min \left( a_1 + \frac{1}{a_2}, a_2 + \frac{1}{a_3}, \dots, a_{n - 1} + \frac{1}{a_n} , a_n + \frac{1}{a_1} \right). \] Prove the inequality \[ \sqrt[n]{a_1 a_2 \dots a_n} + \frac{1}{\sqrt[n]{a_1 a_2 \dots a_n}} \ge m. \]
Let $0\leq a_1\leq a_2\leq \cdots\leq a_{n-1}\leq a_n $ and $a_1+a_2+\cdots+a_n=1.$ Prove that: For any non-negative numbers $x_1,x_2,\cdots,x_n ; y_1, y_2,\cdots, y_n$ , have $$\left(\sum_{i=1}^n a_ix_i - \prod_{i=1}^n x_i^{a_i}\right) \left(\sum_{i=1}^n a_iy_i - \prod_{i=1}^n y_i^{a_i}\right) \leq a_n^2\left(n\sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i} - \sum_{i=1}^n\sqrt{x_i} \sum_{i=1}^n\sqrt{y_i}\right)^2.$$
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 $x_1,\ldots x_n.$
Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1. \] Prove that \[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. \]
Let $a_1, a_2, \ldots, a_n$ be positive real numbers. Prove the inequality \[\binom n2 \sum_{i<j} \frac{1}{a_ia_j} \geq 4 \left( \sum_{i<j} \frac{1}{a_i+a_j} \right)^2\]