A = $2(x-1)^{2}$ + $y^{2}$ + $2021$
Vì $2(x-1)^{2}$≥0∀x
$y^{2}$≥0∀y
=> $2(x-1)^{2}$ + $y^{2}$ ≥0∀x,y
=> $2(x-1)^{2}$ + $y^{2}$ + $2021$≥2021∀x,y
hay A ≥2021∀x,y
Dấu '=' xảy ra
<=> $\left \{ {{2(x-1)^{2} = 0} \atop {y^{2} = 0}} \right.$
<=> \(\left[ \begin{array}{l}2(x-1) = 0\\y = 0\end{array} \right.\)
<=> $\left \{ {{x-1 = 0} \atop {y = 0}} \right.$
<=> $\left \{ {{x=1} \atop {y = 1}} \right.$
Vậy Min A = 2021 <=> x = 1 và y = 0