Đặt A = \(x^2+y^2-xy+y-x+1=x^2-xy-x+y^2+y+1=x^2-2x\frac{y+1}{2}+\frac{y^2+2y+1}{4}-\frac{y^2+2y+1}{4}+y^2+y+1\)
\(=\left(x-\frac{y+1}{2}\right)^2\)\(+\frac{3}{4}y^2\)\(+\frac{1}{2}y\)\(+\frac{3}{4}\)\(=\left(x-\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+2y\frac{1}{3}+\frac{1}{9}+\frac{8}{9}\right)\)
\(=\left(x-\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{1}{3}\right)^2\)\(+\frac{2}{3}\ge\frac{2}{3}\)
Dấu = xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\frac{y+1}{2}\\y=\frac{-1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{-1}{3}\end{matrix}\right.\)
Vậy min A=\(\frac{2}{3}\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{-1}{3}\end{matrix}\right.\)
