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?

Hint
As we know:
$$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$$
and $$x^2+y^2+z^2=2(xy+yz+zx)$$
So, 
$$x^3+y^3+z^3=\frac{1}{2}(x+y+z)(x^2+y^2+z^2)+3xyz$$
Substituting this in the expression gives:
$$\frac{\frac{1}{2}(x+y+z)(x^2+y^2+z^2)+3xyz}{(x+y+z)(x^2+y^2+z^2)}$$
$$=\frac{1}{2}+\frac{3xyz}{(x+y+z)(x^2+y^2+z^2)}$$
Now, substitute $x^2+y^2+z^2=2(xy+yz+zx)$ to get,
$$\frac{1}{2}+\frac{3}{2(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}$$
You should now be able to do this.
