2013 USAJMO Problems/Problem 6
Solution with Thought Process
Without loss of generality, let . Then
.
Suppose x = y = z. Then , so
. It is easily verified that
has no solution in positive numbers greater than 1. Thus,
for x = y = z. We suspect if the inequality always holds.
Let x = 1. Then we have , which simplifies to
and hence
Let us try a few examples: if y = z = 2, we have
; if y = z, we have
, which reduces to
. The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let
! Thus,
, and the claim holds for x = 1.