\(A=x^2+xy+y^2-4x-5y+2021\)
\(4A=4x^2+4xy+4y^2-16x-20y+8084\)
\(=\left(2x+y\right)^2-8\left(2x+y\right)+3y^2-12y+8084\)
\(=\left(2x+y-4\right)^2+3\left(y-2\right)^2+8056\ge8056\)
\(\Rightarrow A\ge2014\)
Đẳng thức xảy ra tại \(x=1;y=2\)
\(A=x^2+xy+y^2-4x-5y+2021\)
\(=\left(x^2+xy-4x\right)+y^2-5y+2021\)
\(=\left[x^2+2.x\left(y-4\right)\frac{1}{2}+\frac{\left(y-4\right)^2}{4}\right]-\frac{\left(y-4\right)^2}{4}+y^2-5y+2021\)
\(=\left(x+\frac{y-4}{2}\right)^2+\frac{3}{4}y^2-3y+2017\)
\(=\left(x+\frac{y-4}{2}\right)^2+\frac{3}{4}\left(y^2-2.y.2+4\right)+2014\)
\(=\left(x+\frac{y-4}{2}\right)^2+\frac{3}{4}\left(y-2\right)^2+2014\ge2014\)
Dấu "=" xảy ra <=> x = 1; y = 2
Vậy min A = 2014 tại x = 1; y =2.