# 2005 IMO Shortlist Problems/A3

## Problem

(*Czech Republic*)
Four real numbers satisfy

and .

Prove that holds for some permutation of .

## Solutions

### Solution 1

**Lemma.** .

*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

.

Hence

.

It follows that at least one of must be at least .

### Solution 2

Without loss of generality, we assume .

**Lemma.** .

*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 . ∎

Now,

,

so , as desired.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*