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.

Scaling, one may assume that $z = -1$, so that $x + y = 1$. Then the desired inequality is
$$6(x^3 + y^3 - 1)^2 \leq (x^2 + y^2 + 1)^3$$
Since $x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1 - 3xy$ and $x^2 + y^2 + 1 = (x + y)^2 - 2xy + 1 = 2 - 2xy$, the desired inequality is
$$54(xy)^2 \leq (2 - 2xy)^3$$
Thus it makes sense to look at $xy = x(1 - x) = {1 \over 4} - (x+{1 \over 2})^2$, whose range is $(-\infty, {1 \over 4}]$. Letting $r = xy$, we need that for $r \leq {1 \over 4}$ we have
$$54r^2 \leq (2 - 2r)^3$$
However, the polynomial $54r^2 - (2 - 2r)^3$ can be directly factorized into $2(r + 2)^2(4r - 1)$, which is nonpositive in the domain $(-\infty, {1 \over 4}]$. Hence the inequality holds.
