Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)=x^2+x(-2y-12)+6y^2+2y+41\ge 0, \forall x \in \Bbb{R}.$$
$$\implies B^2-4AC=(2y+12)^2-4(6y^2+2y+41)=-20(y-1)^2 \le 0,$$
which is true and the equality holds if $y=1$ this further means that $f(x)=(x-7)^2.$
So $f(x,y) \ge 0$, the equality holds if $x=7$ and $y=1$.
The question is: What could be other ways of proving this.
Edit: it will be interesting to note that this quadratic of $x$ and $y$ would represent just a point that is $(7,1)$. It is more clear by the solution of @Aqua given below.

$f(x,y)=x^2-2xy+6y^2-12x+2y+41$
$\frac{\partial f}{dx}=2x-2y-12$
$\frac{\partial f}{dy}=-2x+12y+2$
Stationary points are found when $\frac{\partial f}{dx}=0$ and $\frac{\partial f}{dy}=0$
$2x-2y-12=0$
$-2x+12y+2=0$
Adding the equations gives $10y-10=0\Rightarrow y=1$
Substitute to get $2x-2-12=0\Rightarrow 2x=14\Rightarrow x=7$
Stationary point is at $(7,1)$
Find $\frac{\partial^2 f}{dx^2}=2$ and $\frac{\partial^2 f}{dy^2}=12$ to determine that this is a minimum point (might otherwise have been a maximum or saddle point).
Work out the value $f(7,1)=49-14+6-84+2+41=0$
This is the minimum value of $f(x,y)$.
Therefore $f(x,y)\ge 0$
