2005 IMO Shortlist Problems/A3
(Czech Republic) Four real numbers satisfy
Prove that holds for some permutation of .
Proof. Suppose on the contrary that . Then majorizes , and since is a convex function, by Karamata's Inequality, . But since is strictly convex, equality occurs only when is a permutation of , a contradiction, since we assumed . ∎
Without loss of generality, let . Now, since is increasing on the interval ,
Also, we note
It follows that at least one of must be at least .
Without loss of generality, we assume .
Proof 1. By the Rearrangement Inequality,
As in the first solution, we see , so . It follows that
This is equivalent to
so either or . But since , , so . ∎
Proof 2. From the identity we have
Since , this implies
and since , this gives us , or . Thus
so . ∎
so , as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.