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.

$$z=-(x+y)$$
$$\implies 6 [x^3 + y^3 - (x+y)^3]^2 - [x^2 + y^2 +(x+y)^2]^3$$
$$=-8 x^6 - 24 x^5 y + 6 x^4 y^2 + 52 x^3 y^3 + 6 x^2 y^4 - 24 x y^5 - 8 y^6$$
$$=-2\,(x-y)^2 \, (2 x^2 + 5 x y + 2 y^2)^2\leq 0$$
