\(b,D=x^2+xy+y^2-3x-3y\)
Ta có: \(D+3=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)\)
Đặt: \(\left\{{}\begin{matrix}x-1=a\\y-1=b\end{matrix}\right.\)
Thì \(C+3=a^2+b^2+ab\ge0\left(\forall a,b\right)\)
\(\Rightarrow Min_C=-3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b\Leftrightarrow x=y=1\)