2004 USAMO Problems/Problem 5
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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
.
We present two proofs of this inequality.
First, Hölder's Inequality gives us
![$\begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \\ \qquad \qquad \qquad \displaystyle \left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3 \end{matrix}$](http://latex.artofproblemsolving.com/b/c/9/bc9a514e019732aec002aef72b7a7886bc799e49.png)
.
Setting 
, 
, 
, and 
when 
gives us the desired inequality, with equality when 
.
Second, we may apply the Cauchy-Schwarz Inequality twice to obtain
![$\begin{matrix} \displaystyle \left[(a^2 + 1 + 1)(1 + b^2 + 1)\right] \left[ (1 + 1 + c^2)(a + b + c) \right] & \ge & (a + b + 1)^2( a + b + c^2)^2 \\ & \ge & (a + b + c)^4 \qquad \qquad \quad \; \; \end{matrix}$](http://latex.artofproblemsolving.com/7/a/c/7ac8f2f5d9c8db9a8ddbf0937d7c16af71f0c2aa.png)
,
as desired.
Unfortunately, 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.
