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
. Template:Halmos
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
. Template:Halmos
Proof 2. From the identity
we have
![]()
.
Since
, this implies
and since
, this gives us
, or
. Thus
so
. Template:Halmos
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.