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 expression we want to maximise.
Using the following notation: $S_1=x+y+z$, $S_2=xy+xz+yz$ and $S_3=xyz$. 
From the hypothesis we get that, $S_1^2=4S2$. 
So the expression we want to maximise is:
$P=\dfrac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}=\dfrac{S_1^3-3S_1S_2+3S3}{2S_1S_2}$
Then, simplify it using the hypothesis, in a way such that we only get in terms of $S_1$ and $S_3$ :
$P=\dfrac{1}{2}+\dfrac{6S_3}{S_1^3}$
Now, consider a polynomial with roots $x,y,z$. Ofcourse it is $F(x)=x^3-S_1x^2+S_2x-S_3$. Then, for the roots to be real, the discriminant of this polynomial must be $\geq 0$, so:
$S_1^2S_2^2-4S_2^3-4S_1^3S_3+18S_1S_2S_3-27S_3^2 \geq 0$
Again, using the hypothesis, we get:
$\dfrac{1}{2}S_1^3-27S_3 \geq 0$
Hence, $\dfrac{S_3}{S_1^3} \leq \dfrac{1}{54}$
Finally, $P \leq \dfrac{1}{2}+6\left(\dfrac{1}{54}\right)=\boxed{\dfrac{11}{18}}$
