# Difference between revisions of "2004 JBMO Problems/Problem 1"

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+ | ==Solution 2== | ||

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+ | By Trivial inequality, | ||

+ | <cmath> (x - y)^2 \geq 0 \iff 2(x^2 - xy + y^2) \geq 0 \iff \frac{2}{x^2 + y^2} \geq \frac{1}{x^2 - xy + y^2}.</cmath> | ||

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+ | Then by multiplying by <math>x + y</math> on both sides, we use the Trivial Inequality again to obtain <math>2(x^2 + y^2) \geq (x + y)^2</math> which means <cmath>\frac{x+y}{x^2 - xy + y^2} \leq \frac{2(x + y)}{x^2 + y^2} \leq \frac{2\sqrt{2(x^2+y^2)}}{x^2 + y^2}</cmath> which after simplifying, proves the problem. |

## Revision as of 05:50, 17 June 2021

## Problem

Prove that the inequality holds for all real numbers and , not both equal to 0.

## Solution

Since the inequality is homogeneous, we can assume WLOG that xy = 1.

Now, substituting , we have:

, thus we have

Now squaring both sides of the inequality, we get:
after cross multiplication and simplification we get:
or, which is always true since .

## Solution 2

By Trivial inequality,

Then by multiplying by on both sides, we use the Trivial Inequality again to obtain which means which after simplifying, proves the problem.