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 .
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