2004 USAMO Problems/Problem 5
Problem 5
(Titu Andreescu)
Let 
, 
, and 
be positive real numbers. Prove that
.
Solutions
We first note that for positive 
, 
. We may prove this in the following ways:
- Since
and 
must be both lesser than, both equal to, or both greater than 1, by the rearrangement inequality, 
.
and 
must be both lesser than, both equal to, or both greater than 1, by the 
.- Since
and 
have the same sign, 
, with equality when 
.
and 
have the same sign, 
, with equality when 
.- By weighted AM-GM,
and 
. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of Muirhead's Inequality.
and 
. Adding these gives the desired inequality. Equivalently, the desired inequality is a case of It thus becomes sufficient to prove that
.
This follows from Hölder's Inequality
![$\begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \\ \qquad \qquad \qquad\left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3 \end{matrix}$](http://latex.artofproblemsolving.com/4/c/4/4c42692bbd73cc42b59dfe6dafea52e9cc8a7545.png)
when we take 
, 
, 
, and 
when 
. We have equality if and only if 
.
It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that 
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
