Prove:$x^2+2xy+3y^2-6x-2y\ge-11\;\;\forall(x,y)\in\Bbb{R}$ 
Problem Statement:-
Prove that for all real values of $x$ and $y$
  $$x^2+2xy+3y^2-6x-2y\ge-11$$

I have no idea how to approach this question all I could think on seeing it was tryin to find the linear factors, turns out that the determinant 
$$\begin{vmatrix}
a & h & g\\
h & b & f\\
g & f & c\\
\end{vmatrix}\neq0$$
So finding linear factor just flew out of the window, so I plotted the equation $x^2+2xy+3y^2-6x-2y=0$ turns out it is an ellipse. I could conclude no further as to how to approach this problem.

We have
$$\begin{align}x^2+2xy+3y^2-6x-2y+11&=x^2+(2y-6)x+3y^2-2y+11\\\\&=(x+y-3)^2-(y-3)^2+3y^2-2y+11\\\\&=(x+y-3)^2+2y^2+4y+2\\\\&=(x+y-3)^2+2(y+1)^2\end{align}$$
