<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] Here is an other elementary one.
Since $\displaystyle{\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}}$, we have that: 
$\displaystyle{\frac{xy}{xy+x+y}=\frac{1}{1+\frac{1}{x}+\frac{1}{y}}\leq \frac{1}{1+\frac{4}{x+y}}=1-\frac{4}{4+x+y}}$
Similarly we get the other two inequalities. Thus it suffices to prove that:
$\displaystyle{\frac{6+x^2+y^2+z^2}{9}\geq 1-\frac{4}{4+x+y}+1-\frac{4}{4+y+z}+1-\frac{4}{4+z+x}\Leftrightarrow \\ 5+\frac{x^2+y^2}{2}+5+\frac{y^2+z^2}{2}+5+\frac{z^2+x^2}{2}+\frac{36}{x+y+4}+\frac{36}{y+z+4}+\frac{36}{z+x+4}\geq 36}$
But 
$\displaystyle{\frac{x^2+y^2}{2}+1=\frac{x^2+1}{2}+\frac{y^2+1}{2}\geq x+y}$ and so: 
$\displaystyle{5+\frac{x^2+y^2}{2}+\frac{36}{x+y+4}\geq (5+\frac{x^2+y^2}{2})+\frac{36}{(5+\frac{x^2+y^2}{2})}\geq2\sqrt{(5+\frac{x^2+y^2}{2})\frac{36}{(5+\frac{x^2+y^2}{2})}}=12}$
Simlarly we get the other two inequalities with $y,z$ and $z,x$, we add them and we are done.
Note: No need of AM-GM for $6$ terms.
</s>