Ta có : $x+2y=5$
$\to x = 5-2y$
Có $x^2+4y^2$
$ = (5-2y)^2+4y^2$
$ = 4y^2-20y+25 +4y^2$
$ = 8y^2-20y+25$
$= 8.\bigg(y^2-\dfrac{5}{2}y + \dfrac{25}{8}\bigg)$
$= 8.\bigg(y-\dfrac{5}{4}\bigg)^2+\dfrac{25}{2} ≥ \dfrac{25}{2}$
Dấu "=" xảy ra $⇔y=\dfrac{5}{4}, x= \dfrac{5}{2}$