Help in justifying that $f(a,b)$ is a global minimum Let $A,B,C,D,E,F\in\mathbb{R}$, such that $A>0$ and $B^2<AC$. Given
$f(x,y)= Ax^2+2Bxy +Cy^2 +2Dx + 2Ey + F$
show that it assumes a global minimum:
$f(a,b) = \frac{1}{\delta}\begin{vmatrix}
A & B & D \\
B & C & E \\ 
D & E & F \\ 
\end{vmatrix}$ = $Da+Eb+F$,
where $\delta = 
\begin{vmatrix}
A & B \\
B & C \\
\end{vmatrix}$.
Attempt:
The gradient of $f$ is:
$\nabla f(x,y)= (2Ax + 2By + 2D, 2Bx + 2Cy + 2E)$.
Hence the Hessian of $f$:
$H(x)= \begin{bmatrix}
     2A      & 2B \\
    2B      & 2C \\
\end{bmatrix}$
Just note then that $det(H(x))= 4(AC-B^2)$. Since by hypothesis we have that $A>0$, $0<AC-B^2$, and the Hessian is independent of any point, this shows that $H(x)$ has a global minimum.
Now my problem is: how can i justifiy that the point of minimum is the one given? Should i use Cramer's rule?
Edit: Should i rotate it and transform into a quadratic form of a diagonal matrix? I may guess that the function defines a paraboloid. And supposing that i have found the point of global minimum of the rotated function, how can i plug it back in the original coordinate system and justify that it is the same as $(a,b)$?
Edit: I had some advances. Solving the linear system $\nabla f=0$, using cramer's rule we find that $a = (BE-DC)/\delta$ and $b=(DB-AE)/\delta.$ Now opening the determinant of $f(a,b)$ by cofactors, we find that $f(a,b)$ is exactly equal to $Da+Eb+F.$ This is a great advance, i think. But i don't think that it is totally conving, because i still have to show that pluging in $(a,b)$ in $f$  i shall  get the same value.
 A: 
Edit: I had some advances. Solving the linear system $\nabla f=0$, using cramer's rule we find that $a = (BE-DC)/\delta$ and $b=(DB-AE)/\delta.$ Now opening the determinant of $f(a,b)$ by cofactors, we find that $f(a,b)$ is exactly equal to $Da+Eb+F.$ This is a great advance, i think. But i don't think that it is totally conving, because i still have to show that pluging in $(a,b)$ in $f$  i shall  get the same value.

So you found (with $\delta = AC-B^2$):
$$a = \frac{BE-DC}{\delta} \quad \mbox{and} \quad b = \frac{DB-AE}{\delta}$$
Plugging into:
$$f(x,y)= Ax^2+2Bxy +Cy^2 +2Dx + 2Ey + F$$
gives you:
$$\color{blue}{f(a,b)=} \color{red}{A\left( \tfrac{BE-DC}{\delta} \right)^2+2B\tfrac{BE-DC}{\delta}\tfrac{DB-AE}{\delta} +C\left( \tfrac{DB-AE}{\delta} \right)^2 +D\tfrac{BE-DC}{\delta} + E\tfrac{DB-AE}{\delta}} \\ + \color{blue}{Da+Eb+ F}$$
Rewriting the red part:
$$\frac{1}{\delta^2} \bigl( A(BE-DC)^2+2B(BE-DC)(DB-AE)+C(DB-AE)^2 \\ +\delta D (BE-DC) + \delta E (DB-AE) \bigr)$$
Expanding all and simplifying gives you $0$, leaving you with the blue part for the minimal value.
