2010-2011 Mock USAJMO Problems/Solutions/Problem 2
Problem 2
Let 
be positive real numbers such that 
. Prove that
![\[\frac{3x+1}{y+z}+\frac{3y+1}{z+x}+\frac{3z+1}{x+y}\ge\frac{4}{2x+y+z}+\frac{4}{x+2y+z}+\frac{4}{x+y+2z},\]](http://latex.artofproblemsolving.com/9/f/3/9f38d8c69777a0bdce430ec7ae8fd6be1f6c7561.png)
with equality if and only if 
.
Solution 1
First, we change the terms using the relationship 
:
![\[\sum_{cyc}\frac{3x+1}{y+z} - \sum_{cyc}\frac{4}{2x+y+z}\]](http://latex.artofproblemsolving.com/a/6/e/a6e178aa9a273aa35aceaa065ff1efea1fe4054d.png)
![\[= \sum_{cyc}(\frac{4}{y+z}-1) - \sum_{cyc}\frac{4}{1+x}= 4\sum_{cyc} (\frac{1}{1-x} - \frac{1}{1+x}) - 9\]](http://latex.artofproblemsolving.com/9/1/f/91f911e65f445420bfb3cb646130cc110247f54b.png)
![\[= 8\sum_{cyc} \frac{x}{1-x^2} - 9\]](http://latex.artofproblemsolving.com/e/e/f/eef47264844bf5fb47dff8397893821b6c23d5e4.png)
Then, by Cauchy-Schwarz Inequality, one has:
![\[\sum_{cyc}\frac{x}{1-x^2} \cdot \sum_{cyc} (x-x^3) \ge (x+y+z)^2 = 1\]](http://latex.artofproblemsolving.com/5/6/c/56ccd7de22803d006c442c0e7169c20eee823070.png)
And by AM-GM, one has:
![\[\sum_{cyc} (x-x^3) = 1- \sum_{cyc} x^3 \le 1- 3(\frac{x+y+z}{3})^3 = \frac{8}{9}\]](http://latex.artofproblemsolving.com/f/6/a/f6afbf88739d5a6595bbad3d659ed25c0951cff9.png)
Therefore,
![\[8\sum_{cyc} \frac{x}{1-x^2} - 9 \ge 8 \cdot \frac{1}{\sum_{cyc} (x-x^3)} - 9 \ge 8 \cdot \frac{9}{8} - 9 = 0,\]](http://latex.artofproblemsolving.com/9/f/8/9f850446f4ca1b6171c4a9dcbd9d2cc46ac94fe5.png)
where both equations are true if and only if 
.
edited by User:lightest 17:35, 24 April 2012 (EDT)
