Olympiad problems

2015 Baltic Way, 2

Let $n$ be a positive integer and let $a_1,\cdots ,a_n$ be real numbers satisfying $0\le a_i\le 1$ for $i=1,\cdots ,n.$ Prove the inequality \[(1-{a_i}^n)(1-{a_2}^n)\cdots (1-{a_n}^n)\le (1-a_1a_2\cdots a_n)^n.\]

2019 Balkan MO Shortlist, A4

Tags: inequality , algebra , inequalities , n-variable inequality

Let $a_{ij}, i = 1, 2, \dots, m$ and $j = 1, 2, \dots, n$ be positive real numbers. Prove that \[ \sum_{i = 1}^m \left( \sum_{j = 1}^n \frac{1}{a_{ij}} \right)^{-1} \le \left( \sum_{j = 1}^n \left( \sum_{i = 1}^m a_{ij} \right)^{-1} \right)^{-1} \]

1970 IMO Longlists, 30

Tags: inequality , n-variable inequality

Let $u_1, u_2, \ldots, u_n, v_1, v_2, \ldots, v_n$ be real numbers. Prove that \[1+ \sum_{i=1}^n (u_i+v_i)^2 \leq \frac 43 \Biggr( 1+ \sum_{i=1}^n u_i^2 \Biggl) \Biggr( 1+ \sum_{i=1}^n v_i^2 \Biggl) .\]

2021 Canadian Junior Mathematical Olympiad, 4

Tags: algebra , inequalities , n-variable inequality

Let $n\geq 2$ be some fixed positive integer and suppose that $a_1, a_2,\dots,a_n$ are positive real numbers satisfying $a_1+a_2+\cdots+a_n=2^n-1$. Find the minimum possible value of $$\frac{a_1}{1}+\frac{a_2}{1+a_1}+\frac{a_3}{1+a_1+a_2}+\cdots+\frac{a_n}{1+a_1+a_2+\cdots+a_{n-1}}$$

2017 China Girls Math Olympiad, 5

Tags: inequalities , n-variable inequality

Let $0=x_0<x_1<\cdots<x_n=1$ .Find the largest real number$ C$ such that for any positive integer $ n $ , we have $$\sum_{k=1}^n x^2_k (x_k - x_{k-1})>C$$

2015 Saudi Arabia IMO TST, 3

Tags: inequalities , n-variable inequality , algebra

Let $a_1, a_2, ...,a_n$ be positive real numbers such that $$a_1 + a_2 + ... + a_n = a_1^2 + a_2^2 + ... + a_n^2$$ Prove that $$\sum_{1\le i<j\le n} a_ia_j(1 - a_ia_j) \ge 0$$ Võ Quốc Bá Cẩn.

2019 China Second Round Olympiad, 2

Tags: algebra , combinatorics , inequalities , n-variable inequality

Let $a_1,a_2,\cdots,a_n$ be integers such that $1=a_1\le a_2\le \cdots\le a_{2019}=99$. Find the minimum $f_0$ of the expression $$f=(a_1^2+a_2^2+\cdots+a_{2019}^2)-(a_1a_3+a_2a_4+\cdots+a_{2017}a_{2019}),$$ and determine the number of sequences $(a_1,a_2,\cdots,a_n)$ such that $f=f_0$.

2020 South East Mathematical Olympiad, 4

Tags: inequalities , algebra , n-variable inequality

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 $$\left(\sum_{i=1}^n a_ix_i - \prod_{i=1}^n x_i^{a_i}\right) \left(\sum_{i=1}^n a_iy_i - \prod_{i=1}^n y_i^{a_i}\right) \leq a_n^2\left(n\sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i} - \sum_{i=1}^n\sqrt{x_i} \sum_{i=1}^n\sqrt{y_i}\right)^2.$$

2001 Czech-Polish-Slovak Match, 1

Tags: inequalities , n-variable inequality , holder , convexity

Let $n\ge2$ be a natural number, and $a_i$ be positive numbers, where $i=1,2,\cdots,n.$ Show that \[\left(a_1^3+1\right)\left(a_2^3+1\right)\cdots\left(a_n^3+1\right) \geq \left(a_1^2a_2+1\right)\left(a_2^2a_3+1\right)\cdots\left(a_n^2a_1+1\right)\]

2007 Singapore MO Open, 1

Tags: pigeonhole principle , n-variable inequality , algebra , combinatorics

Let $x_1,x_2,\ldots,x_n$ be real numbers satisfying $x_1^2+x_2^2+\ldots+x_n^2=1$. Prove that for every integer $k\ge2$ there are integers $a_1,a_2,\ldots,a_n$, not all zero, such that $|a_i|\le k-1$ for all $i$, and $|a_1x_1+a_2x_2+\ldots+a_nx_n|\le{(k-1)\sqrt n\over k^n-1}$.

2016 Romania National Olympiad, 2

Tags: inequalities , n-variable inequality

Let be a natural number $ n\ge 2 $ and $ n $ positive real numbers $ a_1,a_n,\ldots ,a_n $ that satisfy the inequalities $$ \sum_{j=1}^i a_j\le a_{i+1} ,\quad \forall i\in\{ 1,2,\ldots ,n-1 \} . $$ Prove that $$ \sum_{k=1}^{n-1} \frac{a_k}{a_{k+1}}\le n/2 . $$

2010 Contests, 2

Tags: n-variable inequality , inequalities

Given the positive real numbers $a_{1},a_{2},\dots,a_{n},$ such that $n>2$ and $a_{1}+a_{2}+\dots+a_{n}=1,$ prove that the inequality \[ \frac{a_{2}\cdot a_{3}\cdot\dots\cdot a_{n}}{a_{1}+n-2}+\frac{a_{1}\cdot a_{3}\cdot\dots\cdot a_{n}}{a_{2}+n-2}+\dots+\frac{a_{1}\cdot a_{2}\cdot\dots\cdot a_{n-1}}{a_{n}+n-2}\leq\frac{1}{\left(n-1\right)^{2}}\] does holds.