bluecarneal wrote: Let $x,y,z$ be positive real numbers. Prove that \[ \sum_{cyclic} \frac{xy}{xy+x^2+y^2} ~\le~ \sum_{cyclic} \frac{x}{2x+z} \] (Proposed by Šefket Arslanagić, Bosnia and Herzegovina) by Cauchy-Schwarz: \[ \frac{x}{2x+z}+\frac{y}{2y+x}+\frac{z}{2z+y}\geq \frac{(x+y+z)^{2}}{2\sum{x^{2}}+\sum{xy}}\] Therefore,it's suffices to prove that: \[ \frac{(x+y+z)^{2}}{2\sum{x^{2}}+\sum{xy}}\geq \frac{xy}{x^{2}+y^{2}+xy}+\frac{yz}{y^{2}+z^{2}+yz}+\frac{xz}{x^{2}+z^{2}+xz}\] Which is: \[ \sum{(x-y)^{2}(\frac{1}{3(x^{2}+xy+y^{2})}-\frac{1}{2(2\sum{x^{2}}+\sum{xy})})}\geq 0 \] Done!

What is MMO?

http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=69&sid=bfe62d41e73afb1a71180194afabf4cc

my solution uses brute force. I noted $ a=x/y,b=y/z,c=z/x $ with $ abc=1 $ then by brute force we delete the denominators and we make the new notation $ v=a+b+c,w=ab+bc+ca $ and I made a sum of positive numbers which will be not smaller than $ 0 $. very nice problem anyway! nice solution,pxchg1200!

See here too http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=317376

a solution by using Carlson's inequality by Carlson's ineq we have $RHS\ge\frac{(\sum\sqrt{xy})^2}{\sum y(2x+z)}=\frac{(\sum\sqrt{xy})^2}{3\sum xy}$ so it suffices to prove $\frac{(\sum\sqrt{xy})^2}{3\sum xy}\ge\sum_{cyc}\frac{xy}{x^2+xy+y^2}$ which is equivalent to $\sum_{cyc}\frac{(x-y)^2}{x^2+xy+y^2}\ge 2\frac{\sum xy-\sum x\sqrt{yz}}{\sum xy}$ or $\sum_{cyclic}(\sqrt x-\sqrt y)^2\frac{[(x+y+2\sqrt{xy})(xy+yz+zx)-z(x^2+xy+y^2)]}{(x^2+xy+y^2)(xy+yz+zx)}\ge 0$ which is trivially true! QED