<s>[INST] Little inequality After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not always true like this:
$4(x^2z + y^2x + z^2y)≥12xyz$ from MA - MG
$x^3 + y^3 + z^3≥x^2y + y^2z+z^2x$
and this remains:
$x^2z + y^2x + z^2y≥x^2y + y^2z+z^2x$ which doesn't always hold? or does it?
 [/INST] Buffalo Way helps here very well.
Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus,
$$\sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3\geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$\sum_{cyc}4x^2z\geq12xyz$$ it's enough to prove that
$$\sum_{cyc}(x^3+2x^2z-3x^2y)\geq0,$$ which we can prove by BW again:
Let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus,
$$\sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3\geq0$$
and we are done!
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