$\dfrac{x^2}{y^2} +\dfrac{y^2}{x^2} + 4 \geq 3\left(\dfrac xy +\dfrac yx\right)\qquad (*)$
$\to \left(\dfrac{x^2}{y^2} +2 + \dfrac{y^2}{x^2}\right) - 3\left(\dfrac xy +\dfrac yx\right) + 2 \geq 0$
$\to \left(\dfrac xy +\dfrac yx\right)^2 - 2\cdot\dfrac32\cdot\left(\dfrac xy +\dfrac yx\right) + \dfrac94 \geq \dfrac14$
$\to \left(\dfrac xy +\dfrac yx -\dfrac32\right)^2 \geq \dfrac14\qquad (**)$
Áp dụng bất đẳng thức $AM-GM$ ta có:
$\dfrac xy +\dfrac yx \geq 2\sqrt{\dfrac xy\cdot\dfrac yx}=2$
$\to \dfrac xy +\dfrac yx -\dfrac32 \geq 2 -\dfrac32 =\dfrac12$
$\to \left(\dfrac xy +\dfrac yx -\dfrac32\right)^2 \geq \dfrac14$
$\to (**)$ đúng
$\to (*)$ đúng
Vậy $\dfrac{x^2}{y^2} +\dfrac{y^2}{x^2} + 4 \geq 3\left(\dfrac xy +\dfrac yx\right)$
