Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$ For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$

I think if we let $x=\tan A;y=\tan B;z=\tan C$, then  
$$P\Leftrightarrow \displaystyle \text {sin}A+\text {sin}B+ \text{sin}C\leq \frac{3}{2}. \hspace{2cm}(1)$$
But I can't prove $(1)$. 

For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3\sqrt3}{2}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{\sqrt{x^2+xy+xz+yz}}\leq\frac{3\sqrt3}{2\sqrt{xy+xz+yz}}$$ or
$$\sum_{cyc}\sqrt{x+y}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}.$$
But by C-S $$\left(\sum_{cyc}\sqrt{x+y}\right)^2\leq(1+1+1)\sum_{cyc}(x+y)=6(x+y+z).$$
Thus, it remains to prove that
$$\sqrt{6(x+y+z)}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}$$ or
$$9(x+y)(x+z)(y+z)\geq8(x+y+z)(xy+xz+yz)$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
Done!
