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 $x,y,z >0$?
 A: By Cauchy–Schwarz inequality with


*

*$u=(x,y,z)$

*$v=(xyz,xyz,xyz)$


we have
$$u\cdot v\le |u|\cdot |v|$$
that is
$$x^2yz+xy^2z+xyz^2 \leq\sqrt3\,xyz \le \frac{1}{3}$$
indeed by AM-GM
$$\frac13=\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{x^2y^2z^2}\implies xyz \le\frac{1}{3\sqrt3}$$
A: You can write the inequality as 
$$
xyz(x+y+z)\leq\frac13. 
$$
On the left-hand-side, using Cauchy-Schwarz, 
$$
xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz. 
$$
Now we get the seemingly easier problem of maximizing $xyz$ under $x^2+y^2+z^2=1$. Here symmetry, common sense, or Lagrange multipliers show that the maximum is achieved when $x=y=z$. Then 
$$
xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz\leq\frac{\sqrt3}{(\sqrt3)^3}=\frac13. 
$$
A: 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.
