\(B=2x^2+2xy+y^2-2x+2y+2016\)
\(=\left(x^2+2xy+y^2+2x+2y+1\right)+\left(x^2-4x+4\right)+2011\)
\(=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(x-2\right)^2+2011\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2+2011\ge2011\forall x;y\)có GTNN là 2011
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)
Vậy \(B_{min}=2011\) tại \(x=2;y=-3\)