Finding maxima of a 3-variable function. Let $x,y,z$ be positive real number satisfy $x+y+z=3$
Find the maximum value of 
$P=\frac{2}{3+xy+yz+zx}+(\frac{xyz}{(x+1)(y+1)(z+1)})^\frac{1}{3}$

We'll prove that $$\frac{2}{3+xy+xz+yz}+\sqrt[3]{\frac{xyz}{(1+x)(1+y)(1+z)}}\leq\frac{5}{6}$$
for all positives $x$, $y$ and $z$ such that $x+y+z=3$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since by Holder $\sqrt[3]{(1+x)(1+y)(1+z)}\geq1+\sqrt[3]{xyz}=1+w$, it's enough to prove that
$$\frac{2u^2}{3u^2+3v^2}+\frac{w}{1+w}\leq\frac{5}{6},$$
which says that it's enough to prove the last inequality for a minimal value of $v^2$.
But $x$, $y$ and $z$ are positive roots of the equation $(X-x)(X-y)(X-z)=0$ or
$3v^2X=-X^3+3uX^2+w^3$,  which says that a line $Y=3v^2X$ 
and a graph of $Y=-X^3+3uX^2+w^3$ have three common points.
Now we see that $v^2$ gets a minimal value, when a line  $Y=3v^2X$ 
is a tangent line to the graph of $Y=-X^3+3uX^2+w^3$,
which happens for equality case of two variables.
Since the last inequality is homogeneous, we can assume $w^3=1$.
Id est, it's enough to prove the last inequality for $y=x$ and $z=\frac{1}{x^2}$, which gives
$$(x-1)^2(38x^7+7x^6-24x^5+32x^4+10x^3-12x^2+2x+1)\geq0,$$
which is obviously true.
For $x=y=z$ we get $P=\frac{5}{6}$, which says that the answer is $\frac{5}{6}$.
Done!
