Đặt \(A=3x^2+y^2-2xy-6x-2y+11\)
\(=3x^2+y^2-2y\left(x+1\right)+\left(x+1\right)^2-\left(x+1\right)^2-6x+11\)
\(=\left(y-x-1\right)^2+3x^2-x^2-2x-1-6x+11\)
\(=\left(x-y+1\right)^2+2x^2-8x+10\)
\(=\left(x-y+1\right)^2+2\left(x^2-4x+5\right)\)
\(=\left(x-y+1\right)^2+2\left(x^2-4x+4+1\right)=\left(x-y+1\right)^2+2\left(x-2\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}x-y+1=0\\ x-2=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=x+1=2+1=3\\ \end{cases}\)