Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$

Question

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.

Answer 1

Let $z=x-y$ then $$f(x,y) = z^2+5y^2-12z-10y+41 $$ $$ = (z-6)^2+5(y-1)^2$$

Answer 2

$f(x,y)=x^2-2xy+6y^2-12x+2y+41$

$\frac{\partial f}{dx}=2x-2y-12$

$\frac{\partial f}{dy}=-2x+12y+2$

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 $f(x,y)\ge 0$

Answer 3

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 $(7,1)$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 6 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - 6 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 1 & - 6 \\ - 1 & 6 & 1 \\ - 6 & 1 & 41 \\ \end{array} \right) $$

Answer 4

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.