# Difference between revisions of "1995 IMO Problems/Problem 2"

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Dividing by 2(ab + bc + ac) and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives | Dividing by 2(ab + bc + ac) and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives | ||

− | <cmath> | + | <cmath>\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},</cmath> |

as desired. | as desired. | ||

## Revision as of 23:18, 21 April 2014

## Contents

## Problem

(*Nazar Agakhanov, Russia*)
Let be positive real numbers such that . Prove that

## Solution

### Solution 1

We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.

### Solution 2

We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.

### Solution 3

Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.

### Solution 3b

Without clever notation: By Cauchy-Schwarz,

Dividing by 2(ab + bc + ac) and noting that by AM-GM gives as desired.

### Solution 4

Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*