<s>[INST] Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :

Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following  inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$

My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.
In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.
Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.
$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$
is what I have got so far, and from here I can’t continue.
Am I on the right direction? Any suggestions or hints will be much appreciated.
 [/INST] The difference
$$
(x^2+y^2+z^2)^3-6(x^3+y^3+z^3)^2
$$
for $x+y+z=0$ can be written
$$
2 (x-y)^2 (y-z)^2 (z-x)^2
$$
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