Finding minimum value of a function of two variables 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, I am trying to change quadratic parts to sum of squares (as the "hint" says).
But I failed to change it.
Also,
Why $f(a,b)=Da+Eb+F$ is at this minimum..? 
 A: Critical points in a two-variable function are the solution of the following equations: $ \frac{\partial f}{\partial x} =0$ and $ \frac{\partial f}{\partial y} =0$.
$ \frac{\partial f}{\partial x} = 2Ax+2By+2D=0$
and
$ \frac{\partial f}{\partial y} = 2Bx+2Cy+2E =0 $
This is a simple case of two equations with two unknowns. If we would like to prove that the function has a minimum by changing quadratic form to sum of squares, we have to get rid of cross term ($xy$). This always can be done by a rotation. Assuming the following coordinate transformation:
$\left[ {\begin{array}{c} x' \\ y' \ \end{array} } \right]=\left[ {\begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \ \end{array} } \right]\left[ {\begin{array}{c} x \\ y \ \end{array} } \right]$
one can write $f(x,y)$ in the new coordinate system as $f(x',y')$ and set the coefficient of $x'y'$-term equal to zero. This gives a value for $\theta$. After that, it is easy to complete squares and find the minimum as complete square should be equal to zero.
A: $f(x,y)=Ax^2+2Bxy+Cy^2+2Dx+2Ey+F$ 
$Af=A^2x^2+2ABxy+ACy^2+2ADx+2AEy+AF=(Ax+By+D)^2+(AC-B^2)y^2+2(AE-BD)y+AF-D^2$
Now, multiply through by $AC-B^2$ and do a complete-the-square on the terms after the first one (the terms that don't involve the variable $x$), and you'll have your sum of squares. 
