Áp dụng BĐT Cosi ta có:
$x^2+\dfrac1{x^2}\ge 2\sqrt{x^2.\dfrac1{x^2}}=2$
$y^2+\dfrac1{y^2}\ge 2\sqrt{y^2.\dfrac1{y^2}}=2$
$\Rightarrow x^2+\dfrac1{x^2}+y^2+\dfrac1{y^2}\ge 2+2=4$
Dấu '=' xảy ra $\Leftrightarrow x=\dfrac1x; y=\dfrac1y\Leftrightarrow x=y=\pm 1$