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

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<cmath> \begin{align*}\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} &\ge \frac{(x + y + z)}{2} \\ &\ge \frac{3}{2} \end{align*},</cmath> | <cmath> \begin{align*}\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} &\ge \frac{(x + y + z)}{2} \\ &\ge \frac{3}{2} \end{align*},</cmath> | ||

as desired. | as desired. | ||

+ | |||

+ | === Solution 5 === | ||

+ | Without clever substitutions, and only AM-GM! | ||

+ | |||

+ | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math> | ||

{{alternate solutions}} | {{alternate solutions}} |

## Revision as of 00:51, 16 January 2018

## 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.

### Solution 5

Without clever substitutions, and only AM-GM!

Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that

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