[tex]\frac{\sqrt{x^2+2y^2}}{z}=\sqrt{\frac{x^2+2y^2}{z^2}}=\sqrt{(\frac{x}{z})^2+\frac{(\sqrt{2}y)^2}{z}}[/tex]
Tương tự rồi cộng lại
[tex]VT\geq \sqrt{(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^2+2(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^2}\\\geq \sqrt{9+2.9}=3\sqrt{3}[/tex]