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 $x,y,z >0$?

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
$$\frac13=\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{x^2y^2z^2}\implies xyz \le\frac{1}{3\sqrt3}$$
