2013 USAMO Problems/Problem 4
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 . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.