Question

Cho các số x, y, z dương. Chmr:

\(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\)

Answer 1

Áp dụng BĐT Cauchy:

\(\frac{x^2}{y^2}+1+\frac{y^2}{z^2}+1+\frac{z^2}{x^2}+1\ge2\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\)

\(\Rightarrow\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}+3\ge\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+3\sqrt[3]{\frac{xyz}{xyz}}=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+3\)

\(\Rightarrow\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\)