1995 IMO Problems/Problem 2
Problem
(Nazar Agakhanov, Russia)
Let be positive real numbers such that
. Prove that
Solution
Solution 1
We make the substitution ,
,
. Then
Since
and
are similarly sorted sequences, it follows from the Rearrangement Inequality that
By the Power Mean Inequality,
Symmetric application of this argument yields
Finally, AM-GM gives us
as desired.
Solution 2
We make the same substitution as in the first solution. We note that in general,
It follows that
and
are similarly sorted sequences. Then by Chebyshev's Inequality,
By AM-GM,
, and by Nesbitt's Inequality,
The desired conclusion follows.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.