Ta có: \(A=4x^2+2y^2+4xy-12x-8y+22\)
\(=4x^2+4xy+y^2-12x-6y+y^2-2y+22\)
\(=\left(2x+y\right)^2-6\left(2x+y\right)+9+y^2-2y+1+12\)
\(=\left(2x+y-3\right)^2+\left(y-1\right)^2+12\ge12\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ 2x+y-3=0\end{cases}\Rightarrow\begin{cases}y=1\\ 2x=-y+3=-1+3=2\end{cases}\Rightarrow\begin{cases}y=1\\ x=1\end{cases}\)


