An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$ 
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.

My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$, then equation becomes
$$y^2 + z^2=2xy$$
This implies that $(y-z)^2=0$.
So this implies that the required value is $$\frac{y^3 + z^3}{(y+z)(y^2 + z^2)}=\frac{1}{2}$$
But this wrong as the correct answer is $\frac{11}{18}$.
What is wrong with my method?

Let $P$ be the given expression we want to maximise.
From the hypothesis, we get $xy+yz+zx=\dfrac{1}{4}(x+y+z)^2$
So, $P=\dfrac{x^3+y^3+z^3}{2(x+y+z)(xy+yz+zx)}=\dfrac{2(x^3+y^3+z^3)}{(x+y+z)^3}$
$\qquad \;\;=2\left[\left(\dfrac{x}{x+y+z}\right)^3+\left(\dfrac{y}{x+y+z}\right)^3+\left(\dfrac{z}{x+y+z}\right)^3\right]$
Let $a=\dfrac{x}{x+y+z}; b=\dfrac{y}{x+y+z}; c=\dfrac{z}{x+y+z}$, we have $\left\{\begin{array}{l}a+b+c=1\\ab+bc+ca=\dfrac{1}{4}\end{array}\right]$
Or $\left\{\begin{array}{l}b+c=1-a\\bc=a^2-a+\dfrac{1}{4}\end{array}\right]$
From the inequality $(b+c)^2\ge 4bc$, we get $0\le a\le\dfrac{2}{3}$.
We have:
$P=2(a^3+b^3+c^3)$
$\quad =2(a^3+(b+c)^3-3bc(b+c)$
$\quad =2\left[a^3+(1-a)^3-3\left(a^2-a+\dfrac{1}{4}\right)(1-a)\right]$
$\quad =6a^3-6a^2+\dfrac{3}{2}a+\dfrac{1}{2}$
$\quad =\dfrac{11}{18}+\dfrac{1}{18}(3a-2)(6a-1)^2\le\dfrac{11}{18}$
So, $\max P=\dfrac{11}{18}\approx {\boxed {0.611}}$.
The equality holds when $(x,y,z)=(4k,k,k)$ or any permutations.
