Áp dụng BĐT Bu-nhi-a-cốp-xki ta có:
$(\frac{x^{2}y}{z}+\frac{y^{2}z}{x}+\frac{z^{2}x}{y})(\frac{x^{2}z}{y}+\frac{y^{2}x}{z}+\frac{z^{2}y}{x})\geq (x^{2}+y^{2}+z^{2})^{2}$
Ta có: $(\frac{x^{2}y}{z}+\frac{y^{2}z}{x}+\frac{z^{2}x}{y})-(\frac{x^{2}z}{y}+\frac{y^{2}x}{z}+\frac{z^{2}y}{x})=\frac{(xy+yz+xz)(x-y)(y-z)(x-z)}{xyz}\geq 0$ vì $x\geq y\geq z> 0$
$\Rightarrow (\frac{x^{2}y}{z}+\frac{y^{2}z}{x}+\frac{z^{2}x}{y})^2\geq (x^{2}+y^{2}+z^{2})^{2}$
$\Rightarrow $ đpcm