Đặt \(A=x^2+y^2+2x+4y+16\)
\(A=\left(x^2+2x+1\right)+\left(y^2+4y+4\right)+11\)
\(A=\left(x+1\right)^2+\left(y+2\right)^2+11\)
Mà \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow A\ge11\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x+1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-2\end{cases}}\)
Vậy \(A_{Min}=11\Leftrightarrow\left(x;y\right)=\left(-1;-2\right)\)