Find maximum value by using AM-GM inequality I have a problem: Find the maximum value of $P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ with $x,y,z>0$.
Is there anyway to solve this problem by using the AM-GM inequality ? Thank for your answer.
The first way, I tried to use AM-GM twice at the two sum in the denominator, but I get the sum of the fractions with square roots of $xy, yz, zx$ at its denominators. 
The other way, I tried to factor the denominator and use both AM-GM and the Schwarz inequality but still get the same with the first way, with higher order 

Another way based on Michael Rozenberg's solution.
If you prove $$\sum_{cyc}\frac{a^3}{a^8+1}\leq\frac{3}{2}$$ you can prove the lemma $$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1}+\frac{1}{c^2+c+1}\ge 1$$ for $abc=1$  and $a;b;c\in R^+$ $(\text{Prove by C-S and} (a=xy/z^2;b=yz/x^2;c=xz/y^2))$
Note that we have:  $$\frac{a^3}{a^8+1}\le \frac{3\left(a^2+1\right)}{4\left(a^4+a^2+1\right)}$$
Or $$-\frac{\left(a-1\right)^2\left(3a^8+6a^7+16a^6+14a^5+16a^4+14a^3+12a^2+6a+3\right)}{4\left(a^2-a+1\right)\left(a^2+a+1\right)\left(a^8+1\right)}\le 0$$
It is simple now. :)
