# Difference between revisions of "2013 USAMO Problems/Problem 4"

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By AM-GM, <cmath>(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}</cmath> | By AM-GM, <cmath>(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}</cmath> | ||

<cmath>ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}</cmath>. Now taking the square root of both sides gives the desired. Equality holds when <math>(a-1)(b-1) = 1</math>. | <cmath>ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}</cmath>. Now taking the square root of both sides gives the desired. Equality holds when <math>(a-1)(b-1) = 1</math>. | ||

+ | {{MAA Notice}} |

## Revision as of 15:40, 3 July 2013

Find all real numbers satisfying

## Solution 1 (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. 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 symmetric (or cyclic) 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 .
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