Ta có:[tex]\sqrt[3]{\frac{1}{3}xy}\leq \frac{x+y+\frac{1}{3}}{3}=\frac{3x+3y+1}{9}\Rightarrow \sqrt[3]{xy}\leq \frac{3x+3y+1}{9}:\sqrt[3]{\frac{1}{3}}=\frac{\sqrt[3]{3}(3x+3y+1)}{9}[/tex]
Tương tự ta được [tex]\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{xz}\leq \frac{\sqrt[3]{3}.(3x+3y+1+3y+3z+1+3z+3x+1)}{9}=\sqrt[3]{3}[/tex]
Dấu "=" xảy ra khi x=y=z=1/3