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.

In the original variables, your polynomial is $(x-y-6)^2 + 5(y-1)^2 $ and so is non-negative, equal to zero only when $y=1$ and then $x=7,$  so $(7,1)$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 - 1 & 1 & 0 \\ 
 - 6 &  - 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
0 & 5 & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  - 1 &  - 6 \\ 
0 & 1 &  - 1 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
1 &  - 1 &  - 6 \\ 
 - 1 & 6 & 1 \\ 
 - 6 & 1 & 41 \\ 
\end{array}
\right) 
$$
