Let $x>0$, $y>0$ and $z>0$ such that $xy+yz+xz=3$. Find a maximize of $$P=\frac{1}{x^2+y^2+1}+\frac{1}{y^2+z^2+1}+\frac{1}{z^2+x^2+1}$$
We need to prove $P\le 1$ with $x=y=z=1$
We have: $$\frac{1}{x^2+y^2+1}\le \frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{y^2}+1\right)\Rightarrow P\le \frac{1}{9}\left(2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+3\right)$$
Or we prove $$\frac{1}{9}\left(2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+3\right)\le 1\Leftrightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le 3$$
WLOG $x\ge y\ge z\Rightarrow \frac{1}{x^2}\le \frac{1}{y^2}\le \frac{1}{z^2};xy\ge yz\ge xz$
By Chebyshev's inequality :
$LHS=\left(xy+yz+xz\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge3\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)\ge3\cdot 3\sqrt[3]{\dfrac{y}{x}\cdot\dfrac{z}{y}\cdot\dfrac{x}{z}}=9$
$\Leftrightarrow9\ge\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)^2\Leftrightarrow3\ge\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}$
Is it right? Pls check for me
By C-S $$\sum_{cyc}\frac{1}{x^2+y^2+1}=\sum_{cyc}\frac{2+z^2}{(x^2+y^2+1)(1+1+z^2)}\leq\sum_{cyc}\frac{2+z^2}{(x+y+z)^2}=$$ $$=\frac{x^2+y^2+z^2+6}{(x+y+z)^2}=\frac{x^2+y^2+z^2+2(xy+xz+yz)}{(x+y+z)^2}=1.$$ The equality occurs for $x=y=z=1$, which says that $1$ is a maximal value.