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?
 A: 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.
A: 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, $P \leq \dfrac{1}{2}+6\left(\dfrac{1}{54}\right)=\boxed{\dfrac{11}{18}}$
A: 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.
A: 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 $(u, v) = (\frac12, 0) \iff x=y=\frac14$.
