<s>[INST] Show $\sum \frac{xy}{xy+x+y} \le \frac{6+x^2+y^2+z^2}{9}$ I've spent some time on this problem but I can't find a method to solve it:

Let $x, y, z>0$, prove that:
  $$\frac{xy}{xy+x+y} + \frac{yz}{yz+y+z} + \frac{zx}{zx+z+x} \le \frac{6+x^2+y^2+z^2}{9}$$

That is how I thought : 
first of all, if we have $a, b, x>0$ and $a\le b$, then $$\frac{a}{a+x}\le\frac{b}{b+x}$$ (which can be simply proved by making the product on diagonal)
So I try to find a $b$ such as $b\ge xy$.
From means inequality (geometric mean and root mean square), we have $$\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}}$$
or $$xy \le \frac{x^2+y^2}{2}$$
So $$\frac{xy}{xy+x+y} \le \frac{\frac{x^2+y^2}{2}}{\frac{x^2+y^2}{2}+x+y}$$
Now, I need to find something which is $\le x+y$ and that's where I've got stuck.
Some hints would be really apreciated! Thanks!
 [/INST] Equivenantly we have to prove that
$$\frac{3xy}{xy+x+y}+ \frac{3yz}{yz+y+z}+ \frac{3zx}{zx+z+x}\leq 2+ \frac{x^2+y^2+z^2}{3}$$
Solution
\begin{align*}  
2+ \frac{x^2+y^2+z^2}{3} &=\sum \frac{x^2+y^2+4}{6} \\    
&\geq \sum \sqrt[6]{x^2y^2} \\    
&=\sum \frac{xy}{\sqrt[3]{(xy)xy}} \\    
&\geq \sum \frac{xy}{\left ( xy+x+y \right )/3}\\   
&=\sum \frac{3xy}{xy+x+y}   
\end{align*}
It only makes use of AM-GM inequality.. :)
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