$4x+y=1\\↔y=1-4x$
Thay $y=1-4x$ vào $4x^2+y^2$
$→4x^2+y^2\\=4x^2+(1-4x^2)\\=4x^2+1-8x+16x^2\\=20x^2-8x+1\\=(2\sqrt 5 x)^2-2.2\sqrt 5 x.\dfrac{2\sqrt 5}{5}+\dfrac{4}{5}+\dfrac{1}{5}\\=\bigg(2\sqrt 5 x-\dfrac{2\sqrt 5}{5}\bigg)^2+\dfrac{1}{5}$
Vì $\bigg(2\sqrt 5x -\dfrac{2\sqrt 5}{5}\bigg)^2\ge 0$
$→\bigg(2\sqrt 5x -\dfrac{2\sqrt 5}{5}\bigg)^2+\dfrac{1}{5}\ge \dfrac{1}{5}$
$→$ Dấu "=" xảy ra khi $2\sqrt 5 x-\dfrac{2\sqrt 5}{5}=0$
$↔2\sqrt 5 x=\dfrac{2\sqrt 5}{5}\\↔x=\dfrac{1}{5}\\→y=\dfrac{1}{5}$
Vậy BĐT được chứng minh và dấu "=" xảy ra khi $x=y=\dfrac{1}{5}$
