Problem: (2000 IMO – #2) Let be positive real numbers so that . Prove that
Solution: The standard trick to solve the condition is to substitute
for positive reals (their existence is guaranteed since it’s a three-variable system with three equations).
Our inequality reduces to
Multiplying through by , expanding, and rearranging everything to the RHS, we have
which is just Schur’s Inequality, so the result is proved. QED.
Comment: The condition almost always calls for the substitution above. It also works with in which case you simply have another variable multiplied by each of the terms.
Practice Problem: Go take the 2006 Mock AIME 5 below.
Leave a Reply
You must be logged in to post a comment.