Dễ cm được: \(\frac{x}{y}+\frac{y}{x}\ge2\)
Ta có:
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}-3\left(\frac{x}{y}+\frac{y}{x}\right)+2+3\)
\(=\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}-2\right)+3\ge3\) (Do \(\frac{x}{y}+\frac{y}{x}\ge2\))
Vậy min bằng 3. "=" khi x=y
Ta có:
\(\frac{x^2}{y^2}+2+\frac{y^2}{x^2}=\left(\frac{x}{y}\right)^2+2\cdot\frac{x}{y}\cdot\frac{y}{x}+\left(\frac{y}{x}\right)^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2\)
Theo đó thì: (đặt A cho dễ)
\(A=\frac{x^2}{y^2}+\frac{y^2}{x^2}-3\left(\frac{x}{y}+\frac{y}{x}\right)+5=\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+3\)
Đặt \(t=\frac{x}{y}+\frac{y}{x}\Rightarrow t\ge2\)
\(A=t^2-3t+3=\left(t^2-2\cdot t\cdot\frac{3}{2}+\frac{9}{4}\right)+\frac{3}{4}\)
\(=\left(t-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(A_{min}=\frac{3}{4}\) khi \(t=\frac{3}{2}\Leftrightarrow\frac{x}{y}+\frac{y}{x}=\frac{3}{2}\)
