Find Min Values Of $P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$ Given $x,y,z>0$ and $y+z=x(y^2+z^2)$
Find Min Values Of
$P=\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{4}{(1+x)(1+y)(1+z)}$
Could Someone give me an idea ?

by the Lagrange Multiplier Method we get
$$P\geq \frac{91}{108}$$ the equal sign holds if $$x=\frac{1}{5},y=z=5$$
