How to prove the existence of a minimum of a quadratic function of two variables?

Question

I am given function

$$ f(x,y)=Ax^2+2Bxy+Cy^2+2Dx+2Ey+F,\quad\text{where }A>0\text{ and }B^2<AC . $$

Prove that a point $(a,b)$ exists which $f$ has a minimum.

I figured out that there is no stationary point for this equation.

So, Hessian Matrix seems not helpful.

In my book, it says that "change quadratic part to sum of squares

but, Can't think of any way to change it to sum of squares.

Also,

Why $f(a,b)=Da+Eb+F$ is at this minimum..?

Answer 1

You figured out the wrong thing, since by computing the partial derivatives we have that stationary points occur for: $$\left(\begin{array}{cc} A& B\\ B&C\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) = -\left(\begin{array}{c} D\\ E\end{array}\right).$$ Since the determinant of the matrix on the LHS is positive and $A>0$, there is a unique stationary point that is a global minimum for $f$ - it is the center of the associated conic section.

Answer 2

Let $H= \pmatrix{A & B \\ B & C}, g=\pmatrix{D \\ E}$, then we have $f(x) = \langle x, Hx \rangle + 2 \langle g, x \rangle +F$.

Note that $\det H = AC-B^2 >0$, so $H$ is invertible.

Show that $H$ is positive definite (this is straightforward, but not immediate, use the fact that the eigenvalues of a symmetric real matrix are real). This shows that $f$ is convex, hence a stationary point is a minimiser.

Then ${\partial f(x) \over \partial x}(\delta) = 2(Hx+g)^T \delta$. It is straightforward to find an $x$ such that ${\partial f(x) \over \partial x} = 0$.

Answer 3

Hint. Since you are asked to do it by algebra rather than calculus: $$f(x,y)=\frac{(Ax+By+D)^2}{A}+\frac{((AC-B^2)y+(AE-BD))^2}{A(AC-B^2)} +\frac{(AC-B^2)(AF-D^2)-(AE-BD)^2}{A(AC-B^2)}\ ,$$ which is a sum of two squares plus a constant.