\(A=\frac{1}{4}\left(4x^2+4y^2+4xy-12x-12y\right)+2006\)
\(A=\frac{1}{4}\left(x^2+4y^2+9+4xy-6x-12y\right)+\frac{3}{4}\left(x^2-2x+1\right)+2003\)
\(A=\frac{1}{4}\left(x+2y-3\right)^2+\frac{3}{4}\left(x-1\right)^2+2003\ge2003\)
\(\Rightarrow A_{min}=2003\) khi \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)