Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$

$$z=-(x+y)$$

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

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 $$ 2 (x-y)^2 (y-z)^2 (z-x)^2 $$

Now, let $x^2+y^2=2uxy.$

Thus, since $xy\geq0$ and for $xy=0$ our inequality is true, we can assume that $xy>0$, which gives $u\geq1$ and we need to prove that: $$8(x^2+xy+y^2)^3\geq6(-3xy(x+y))^2$$ or $$2(2u+1)^3\geq27(u+1)$$ or $$16u^3+24u^2-15u-25\geq0,$$ which is obvious for $u\geq1$.