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

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

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

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

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

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 $f(x,y)\ge 0$

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 $(7,1)$

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