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: 787

1986 IMO Longlists, 45

Given $n$ real numbers $a_1 \leq a_2 \leq \cdots \leq a_n$, define \[M_1=\frac 1n \sum_{i=1}^{n} a_i , \quad M_2=\frac{2}{n(n-1)} \sum_{1 \leq i<j \leq n} a_ia_j, \quad Q=\sqrt{M_1^2-M_2}\] Prove that \[a_1 \leq M_1 - Q \leq M_1 + Q \leq a_n\] and that equality holds if and only if $a_1 = a_2 = \cdots = a_n.$

2009 Canada National Olympiad, 3

Define $f(x,y,z)=\frac{(xy+yz+zx)(x+y+z)}{(x+y)(y+z)(z+x)}$. Determine the set of real numbers $r$ for which there exists a triplet of positive real numbers satisfying $f(x,y,z)=r$.

2005 Taiwan National Olympiad, 3

$a_1, a_2, ..., a_{95}$ are positive reals. Show that $\displaystyle \sum_{k=1}^{95}{a_k} \le 94+ \prod_{k=1}^{95}{\max{\{1,a_k\}}}$

2004 Polish MO Finals, 4

Let real numbers $ a,b,c$. Prove that $ \sqrt{2(a^2\plus{}b^2)}\plus{}\sqrt{2(b^2\plus{}c^2)}\plus{}\sqrt{2(c^2\plus{}a^2)}\ge \sqrt{3(a\plus{}b)^2\plus{}3(b\plus{}c)^2\plus{}3(c\plus{}a)^2}$.

1999 Balkan MO, 4

Let $\{a_n\}_{n\geq 0}$ be a non-decreasing, unbounded sequence of non-negative integers with $a_0=0$. Let the number of members of the sequence not exceeding $n$ be $b_n$. Prove that \[ (a_0 + a_1 + \cdots + a_m)( b_0 + b_1 + \cdots + b_n ) \geq (m + 1)(n + 1). \]

2005 Junior Balkan Team Selection Tests - Romania, 14

Let $a,b,c$ be three positive real numbers with $a+b+c=3$. Prove that \[ (3-2a)(3-2b)(3-2c) \leq a^2b^2c^2 . \] [i]Robert Szasz[/i]

1999 Romania Team Selection Test, 5

Let $x_1,x_2,\ldots,x_n$ be distinct positive integers. Prove that \[ x_1^2+x_2^2 + \cdots + x_n^2 \geq \frac {2n+1}3 ( x_1+x_2+\cdots + x_n). \] [i]Laurentiu Panaitopol[/i]

2008 Irish Math Olympiad, 2

For positive real numbers $ a$, $ b$, $ c$ and $ d$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$ prove that $ a^2b^2cd \plus{} \plus{}ab^2c^2d \plus{} abc^2d^2 \plus{} a^2bcd^2 \plus{} a^2bc^2d \plus{} ab^2cd^2 \le 3/32,$ and determine the cases of equality.

2010 Turkey Junior National Olympiad, 4

Prove that \[ a^2b^2(a^2+b^2-2) \geq (a+b)(ab-1) \] for all positive real numbers $a$ and $b.$

2012 Balkan MO Shortlist, A6

Let $k$ be a positive integer. Find the maximum value of \[a^{3k-1}b+b^{3k-1}c+c^{3k-1}a+k^2a^kb^kc^k,\] where $a$, $b$, $c$ are non-negative reals such that $a+b+c=3k$.

2007 Ukraine Team Selection Test, 8

$ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4}\plus{}a_{1}x^{3}\plus{}a_{2}x^{2}\plus{}a_{1}x^{1}\plus{}a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac{3}{10}(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}< M\minus{}m < 3(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}$

2001 Vietnam Team Selection Test, 1

Let’s consider the real numbers $a, b, c$ satisfying the condition \[21 \cdot a \cdot b + 2 \cdot b \cdot c + 8 \cdot c \cdot a \leq 12.\] Find the minimal value of the expression \[P(a, b, c) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.\]

2006 Croatia Team Selection Test, 2

Assume that $a, b,$ and $c$ are positive real numbers for which $(a+b)(a+c)(b+c) = 1$. Prove that $ab+bc+ca \leq\frac{3 }{4}.$

2009 Spain Mathematical Olympiad, 5

Let, $ a,b,c$ real positive numbers with $ abc \equal{} 1$ Prove: $ (\frac {a}{1 \plus{} ab})^2 \plus{} (\frac {b}{1 \plus{} bc})^2 \plus{} (\frac {c}{1 \plus{} ca})^2\geq \frac {3}{4}$ Thanks!

2024 Baltic Way, 4

Find the largest real number $\alpha$ such that, for all non-negative real numbers $x$, $y$ and $z$, the following inequality holds: \[ (x+y+z)^3 + \alpha (x^2z + y^2x + z^2y) \geq \alpha (x^2y + y^2z + z^2x). \]

2011 ELMO Shortlist, 5

Given positive reals $x,y,z$ such that $xy+yz+zx=1$, show that \[\sum_{\text{cyc}}\sqrt{(xy+kx+ky)(xz+kx+kz)}\ge k^2,\]where $k=2+\sqrt{3}$. [i]Victor Wang.[/i]

2016 India National Olympiad, P2

For positive real numbers $a,b,c$ which of the following statements necessarily implies $a=b=c$: (I) $a(b^3+c^3)=b(c^3+a^3)=c(a^3+b^3)$, (II) $a(a^3+b^3)=b(b^3+c^3)=c(c^3+a^3)$ ? Justify your answer.

1997 Romania National Olympiad, 2

I found this inequality in "Topics in Inequalities" (I 85) For all positive reals $x,y,z$ with $xyz=1$ prove: \[ \frac{x^9+y^9}{x^6+x^3y^3+y^6}+\frac{y^9+z^9}{y^6+y^3z^3+z^6}+\frac{z^9+x^9}{z^6+z^3x^3+x^6}\geq 2 \]

2002 Iran MO (3rd Round), 1

Let $a,b,c\in\mathbb R^{n}, a+b+c=0$ and $\lambda>0$. Prove that \[\prod_{cycle}\frac{|a|+|b|+(2\lambda+1)|c|}{|a|+|b|+|c|}\geq(2\lambda+3)^{3}\]

2014 China Western Mathematical Olympiad, 7

In the plane, Point $ O$ is the center of the equilateral triangle $ABC$ , Points $P,Q$ such that $\overrightarrow{OQ}=2\overrightarrow{PO}$. Prove that\[|PA|+|PB|+|PC|\le |QA|+|QB|+|QC|.\]

2010 Regional Competition For Advanced Students, 1

Let $0 \le a$, $b \le 1$ be real numbers. Prove the following inequality: \[\sqrt{a^3b^3}+ \sqrt{(1-a^2)(1-ab)(1-b^2)} \le 1.\] [i](41th Austrian Mathematical Olympiad, regional competition, problem 1)[/i]

2006 France Team Selection Test, 2

Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \] When is there equality?

2007 Kyiv Mathematical Festival, 4

Let $a,b,c>0$ and $abc\ge1.$ Prove that a) $\left(a+\frac{1}{a+1}\right)\left(b+\frac{1}{b+1}\right) \left(c+\frac{1}{c+1}\right)\ge\frac{27}{8}.$ b)$27(a^{3}+a^{2}+a+1)(b^{3}+b^{2}+b+1)(c^{3}+c^{2}+c+1)\ge$ $\ge 64(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1).$ [hide="Generalization"]$n^{3}(a^{n}+\ldots+a+1)(b^{n}+\ldots+b+1)(c^{n}+\ldots+c+1)\ge$ $\ge (n+1)^{3}(a^{n-1}+\ldots+a+1)(b^{n-1}+\ldots+b+1)(c^{n-1}+\ldots+c+1),\ n\ge1.$ [/hide]

2014 China Northern MO, 2

Define a positive number sequence sequence $\{a_n\}$ by \[a_{1}=1,(n^2+1)a^2_{n-1}=(n-1)^2a^2_{n}.\]Prove that\[\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots +\frac{1}{a^2_n}\le 1+\sqrt{1-\frac{1}{a^2_n}} .\]

2006 Switzerland Team Selection Test, 1

Let $a,b,c \in \mathbb{R^+}$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Show $\sqrt{ab+c} + \sqrt{bc+a} + \sqrt{ca+b} \ge \sqrt{abc} + \sqrt{a} + \sqrt{b} + \sqrt{c}$. :D