# 2013 USAMO Problems/Problem 4

Find all real numbers satisfying

## Solution (Cauchy or AM-GM)

The key Lemma is: for all . Equality holds when .

This is proven easily. by Cauchy.

Equality then holds when .

Now assume that . Now note that, by the Lemma,

. So equality must hold in order for the condition in the problem statement to be met. So and . If we let , then we can easily compute that . Now it remains to check that .

But by easy computations, , which is obvious. Also , which is obvious, since .

So all solutions are of the form , and all permutations for .

**Remark:** An alternative proof of the key Lemma is the following:
By AM-GM,
. Now taking the square root of both sides gives the desired. Equality holds when .

## Solution 2

WLOG, assume that . Let and . Then , and . The equation becomes Rearranging the terms, we have Therefore and Express and in terms of , we have and Easy to check that is the smallest among , and Then , and Let , we have the solutions for as follows: and permutations for all

--J.Z.

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