<s>[INST] How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$? How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?
 [/INST] It's true even for all reals $x$, $y$ and $z$.
Indeed, we need to prove that
$$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or
$$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or
$$\sum_{cyc}(2x^4+4x^2y^2-6x^2yz)\geq0$$ or
$$\sum_{cyc}(x^4-2x^2y^2+y^4+3z^2x^2-6z^2xy+3z^2y^2)\geq0$$ or
$$\sum_{cyc}(x-y)^2((x+y)^2+3z^2)\geq0,$$
which is obvious.
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