Find the maximun value of the expression $P=\sum \sqrt[3]{\frac{a^{2}+a}{a^{2}+a+1}}$ Let $a,b,c$ be positive real numbers such that $abc\leq 1$ .Find the maximun value of the expression
$P=\sqrt[3]{\frac{a^{2}+a}{a^{2}+a+1}}+\sqrt[3]{\frac{b^{2}+b}{b^{2}+b+1}}+\sqrt[3]{\frac{c^{2}+c}{c^{2}+c+1}}$

Holder's inequality or Power Means will give
$$ \sum_{cyc} \frac{a^2+a}{a^2+a+1}  \ge \frac19\left(\sum_{cyc} \sqrt[3]{\frac{a^2+a}{a^2+a+1}} \right)^3$$
with equality iff $a=b=c$. We claim that the LHS reaches a maximum of $2$ exactly when $a=b=c=1$, and show the same below.
This is equivalent to showing
$$ \sum_{cyc} \left(1- \frac1{a^2+a+1} \right)  \le 2 \iff \sum_{cyc} \frac1{a^2+a+1} \ge 1 $$
As $abc=1$ for the minimum, there exists positive reals $x, y, z$ s.t. $a = \frac{yz}{x^2}, b = \frac{zx}{y^2}, c = \frac{xy}{z^2}$.  So we can homogenise and rewrite as
$$\sum_{cyc} \frac{x^4}{x^4+x^2yz+y^2z^2} \ge 1$$
Using Cauchy Schwarz on the LHS, it is sufficient to show
$$(x^2+y^2+z^2)^2 \ge \sum_{cyc} x^4 + xyz (x+y+z)+\sum_{cyc} x^2y^2$$
$$\iff \sum_{cyc} x^2y^2 \ge xyz \sum_{cyc} x \iff \sum_{cyc} x^2(y-z)^2 \ge 0$$
Equality holds iff $x=y=z$, i.e. when $a=b=c=1$, so that is when we have the maximum.
