[tex](1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=2+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}[/tex]
Áp dụng BĐT Cauchy ta có:
[tex]\frac{x}{y}+\frac{x}{z}+\frac{x}{x}\geq 3\sqrt[3]{\frac{x^3}{xyz}}=\frac{3x}{\sqrt[3]{xyz}}\Rightarrow \frac{x}{y}+\frac{x}{z}\geq \frac{3x}{\sqrt[3]{xyz}}-1[/tex]
Cộng vế theo vế ta có: [tex](1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=2+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}\geq 2+\frac{3x}{\sqrt[3]{xyz}}+\frac{3y}{\sqrt[3]{xyz}}+\frac{3z}{\sqrt[3]{xyz}}-3=2+2.\frac{x+y+z}{\sqrt[3]{xyz}}+\frac{x+y+z}{\sqrt[3]{xyz}}-3\geq 2+2.\frac{x+y+z}{\sqrt[3]{xyz}}+\frac{3\sqrt[3]{xyz}}{3}-3=2+2.\frac{x+y+z}{\sqrt[3]{xyz}}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
