Ta có
$\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2$
$= \dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 2$
$\Rightarrow \dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 4 = \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 + 2$
Áp dụng bất đẳng thức $BCS$ ta được:
$\dfrac{x^2}{y^2}\cdot\dfrac{y^2}{x^2} \geq \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2$
$\Leftrightarrow 1 \geq \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2$
$\Leftrightarrow 2 \geq 2\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2$
$\Leftrightarrow \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 + 2 \geq \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 + 2\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 = 3\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2$
Vậy $\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} + 4 \geq 3\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2$
Dấu = xảy ra $\Leftrightarrow x = \pm y$
