Ta có $S^{2}=\Sigma \frac{x^{2}y^{2}}{z^{2}}+2(\Sigma x^{2})$ÁD BĐT AM-GM:
$\frac{x^{2}y^{2}}{z^{2}}+\frac{y^{2}z^{2}}{x^{2}}\geq 2y^{2}$
TT$\Rightarrow \Sigma \frac{x^{2}y^{2}}{z^{2}}\geq\Sigma x^{2}$
$\Rightarrow S^{2} \geq 3(x^{2}+y^{2}+z^{2})=3$$\Rightarrow S\geq \sqrt{3}$
Dấu''='' xra$\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}$