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?

Suppose one of the variables is zero.  Then you have shown the maximum is $\frac12$.
So let us consider the case when none of the variables are zero.  Then as both the objective and constraint are symmetric & homogeneous, we may set WLOG $z=1$ and $1\ge x\ge y>0$.  Thus in this case we have the constraint $x^2+y^2+1 = 2(x+y+xy) \iff (x-y)^2+1=2(x+y)$ and need to maximize
$$F=\frac{x^3+y^3+1}{(x+y+1)(x^2+y^2+1)} $$
Now let $u= x+y, v = x-y \implies v^2+1=2u$.  Now we can express everything in terms of $v \in [0, 1)$,
$$F = \frac{(v^2+3)^3+6(v^2-1)^2}{2(v^2+3)^3}=\frac12+\frac3{v^2+3}\left(\frac4{v^2+3}-1\right)^2\le \frac12+\frac3{3}\left(\frac43-1\right)^2=\frac{11}{18}$$
clearly this is a higher maximum than the earlier case, and equality is possible when $(u, v) = (\frac12, 0) \iff x=y=\frac14$.
