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.
 A: Let $z=x-y$ then $$f(x,y) = z^2+5y^2-12z-10y+41 $$ $$ = (z-6)^2+5(y-1)^2$$
A: $f(x,y)=x^2-2xy+6y^2-12x+2y+41$
$\frac{\partial f}{dx}=2x-2y-12$
$\frac{\partial f}{dy}=-2x+12y+2$
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 $f(x,y)\ge 0$
A: 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 $(7,1)$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 - 1 & 1 & 0 \\ 
 - 6 &  - 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
0 & 5 & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  - 1 &  - 6 \\ 
0 & 1 &  - 1 \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
1 &  - 1 &  - 6 \\ 
 - 1 & 6 & 1 \\ 
 - 6 & 1 & 41 \\ 
\end{array}
\right) 
$$
A: We can write given Polynomial in Quadratic equation in (y-1) as-
f(x,y)=$x^2−2xy+6y^2−12x+2y+41$
=$x^2−2xy+y^2+36−12x+12y+5y^2-10y+5$
=${(x-y)}^2$ + ${6}^2$ -2.6.(x-y) + 5${(y-1)}^2$
=${(x-y-6)}^2$ + 5${(y-1)}^2$
=${[(x-7)-(y-1)]}^2$ + 5${(y-1)}^2$
=6${(y-1)}^2$ - 2(x-7)(y-1) + ${(x-7)}^2$
which is Quadratic equation in (y-1).
Now $B^2$-4AC=${[-2(x-7)]}^2$ - 4.6.${(x-7)}^2$
=-20.${(x-7)}^2$ <0 or 0 (at x=7)
= -ve definite or 0,
which implies that f(x,y)=$x^2−2xy+6y^2−12x+2y+41$>0 or 0 i.e. f(x,y)is positive definite & the equality holds if x=7 & y=1 [as it is quadratic also in (y-1)].
Hence proved.
