Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$ 
For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$


We have: $P=\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{\left(z+1\right)^2\left(z^2+1\right)}+\frac{1}{\left(y+1\right)^2\left(y^2+1\right)}+\frac{1}{\left(x+1\right)^2\left(x^2+1\right)}\right)$
$\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{2\left(z^2+1\right)^2}+\frac{1}{2\left(x^2+1\right)^2}+\frac{1}{2\left(y^2+1\right)^2}\right)$
$\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{9}{2\left(\left(z^2+1\right)^2+\left(y^2+1\right)^2+\left(x^2+1\right)^2\right)}\right)$
I can't continue. Help
 A: Let $x=y=z=1$.
Hence, $P=24$.
We'll prove that it's a minimal value.
Indeed, by C-S
$$\sum_{cyc}\frac{(x+1)^2(y+1)^2}{z^2+1}=\sum_{cyc}\frac{(x+1)^2(y+1)^2(x+y)^2}{(z^2+1)(x+y)^2}\geq\frac{\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2}{\sum\limits_{cyc}(z^2+1)(x+y)^2}.$$
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2\geq24\sum\limits_{cyc}(z^2+1)(x+y)^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $u=1$ and$$\sum\limits_{cyc}(x+1)(y+1)(x+y)=\sum_{cyc}(x^2y+x^2z+2x^2+2xy+2x)=$$
$$=9uv^2-3w^3+2u(9u^2-6v^2)+6uv^2+6u^3=3(8u^3+uv^2-w^3);$$
$$\sum\limits_{cyc}(z^2+1)(x+y)^2=2\sum_{cyc}(x^2y^2+x^2yz+x^2u^2+xyu^2)=$$
$$=2(9v^4-6uw^3+3uw^3+9u^4-6u^2v^2+3u^2v^2)=6(3u^4-u^2v^2+3v^4-uw^3).$$
Id est, it's enough to prove that $f(w^3)\geq0$, where
$$f(w^3)=(8u^3+uv^2-w^3)^2-16(3u^6-u^4v^2+3u^2v^4-u^3w^3).$$
Now, $$f'(w^3)=-2(8u^3+uv^2-w^3)+16u^3=2w^3-2uv^2\leq0,$$
which says that $f$ is decreasing function.
Thus, it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$ and $z=3-2x$.
Hence, $$\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2\geq24\sum\limits_{cyc}(z^2+1)(x+y)^2$$
gives
$$(x-1)^2(x^4-2x^3-11x^2+24x+4)\geq0.$$
which is obvious.
Done!
