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.

Thanks for all the answers you guys provided, and after spending a large amount of time wandering in this question and understanding these answers, I came up with my solution which I think is valid.
By AM-GM:
$x^2+xy+y^2\\=\frac{x(x+y)}{2}+\frac{y(x+y)}{2}+\frac{x^2+y^2}{2}\\ \geq3\cdot \sqrt[3]{\frac{xy(x+y)^2}{4}\cdot\frac{x^2+y^2}{2}}\\\geq 3\cdot \sqrt[3]{\frac{x^2y^2z^2}{4}}$
Therefore $(x^2+y^2+z^2)^3\geq8\cdot(3\cdot \sqrt[3]{\frac{x^2y^2z^2}{4}} )^3=54x^2y^2z^2$
Now, since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ and $x+y+z=0$,
we have $6(x^3+y^3+z^3)^2=6(3xyz)^2=54x^2y^2z^2$
Hence the inequality is proven.
