Ta có $4x+y=1$
$\to y = 1-4x$
Ta có : $4x^2+y^2$
$ = 4x^2+(1-4x)^2$
$ = 20x^2-8y+1$
$ 20.\bigg(x^2-\dfrac{2}{5}x+\dfrac{1}{20}\bigg)$
$ = 20.\bigg(x^2-2.x.\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{100}\bigg)$
$ = 20.\bigg(x-\dfrac{1}{5}\bigg)^2 +\dfrac{1}{5} ≥ \dfrac{1}{5}$
Dấu "=" xảy ra $⇔x=y=\dfrac{1}{5}$