Prove inequality $a^3b^3+2b^3c^3+3a^3c^3\le 0$ For $a^3+b^3+c^3=0$. Prove that $a^3b^3+2b^3c^3+3a^3c^3\le 0$

i think from $a^3+b^3+c^3=0$ we have one of the three numbers is zero, the other two are opposites.Suppose $a;b$ opposites and $c=0$

For simplicity, let $a^3=x,b^3=y,c^3=z$. Then: $x+y+z=0\Rightarrow x=-(y+z).$
$$xy+2yz+3xz=$$
$$-y(y+z)+2yz-3z(y+z)=$$
$$-y^2-2yz-3z^2=$$
$$-(y+z)^2-2z^2\le0.$$
