<s>[INST] 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.
 [/INST] 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.
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