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

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=== Solution 3b === | === Solution 3b === | ||

Without clever notation: | Without clever notation: | ||

− | By Cauchy-Schwarz, <cmath>(a(b+c) + b(c+a) + c(a+b)) \cdot (\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)})</cmath> | + | By Cauchy-Schwarz, <cmath>\left(a(b+c) + b(c+a) + c(a+b)\right) \cdot \left(\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)}\right)</cmath> |

− | <cmath>\ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2</cmath> | + | <cmath>\ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2</cmath> |

<cmath>= (ab + bc + ac)^2</cmath> | <cmath>= (ab + bc + ac)^2</cmath> | ||

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

<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> | <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 09:27, 15 July 2017

## 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 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.*