Inequality involving partial derivatives Suppose $f(x, y)$ is a twice continuously differentiable function with a unique minimum at $(0, 0)$. Why at $(0, 0)$ must we have $$\frac{\partial^{2}f}{\partial x^{2}}\frac{\partial^{2}f}{\partial y^{2}} \geq \left(\frac{\partial^{2}f}{\partial x \partial y}\right)^{2}?$$
 A: This kind of complements Robert's answer so that the whole thing is more self contained. Consider the line
$$
\begin{pmatrix}x(t)\\y(t)\end{pmatrix} 
=
\begin{pmatrix}\alpha t\\\beta t\end{pmatrix},
$$
where $\alpha$ and $\beta$ are fixed parameters, and $t$ is a real variable. Then the first and second derivatives of $f$ along this line with respect to $t$ are given by
$$
\frac{\mathrm{d} f}{\mathrm{d}t} = \alpha \, f_x + \beta\,f_y,
\qquad
\frac{\mathrm{d}^2 f}{\mathrm{d}t^2} = \alpha^2 \, f_{xx} + 2\alpha\beta\,f_{xy} + \beta^2\,f_{yy}.
$$
At the minimum, we must have $\frac{\mathrm{d}^2 f}{\mathrm{d}t^2}\geq0$.
In other words,
$$
\begin{pmatrix}\alpha\\\beta\end{pmatrix}^T \begin{pmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{pmatrix}
\begin{pmatrix}\alpha\\\beta\end{pmatrix}
\geq0,
$$
for any $\alpha$ and $\beta$. In particular, we have $f_{xx}\geq0$ and $f_{yy}\geq0$. Let us define
$$
H=\begin{pmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{pmatrix},
$$
and ask if there is $\lambda$ such that $H-\lambda I$ is singular. This leads to the characteristic equation
$$
(f_{xx}-\lambda)(f_{yy}-\lambda)-f_{xy}^2=0,
$$
which has two real solutions because the discriminant is nonnegative:
$$
(f_{xx}+f_{yy})^2 + 4f_{xx}f_{yy}f_{xy}^2\geq0.
$$
Supposing that $\lambda$ assumes one of these two values, we can choose nontrivial vector $\xi=(\alpha,\beta)^T$ from the null space of $H-\lambda I$, meaning that
$$
\xi^TH\xi = \lambda \xi^T \xi \geq0.
$$
This shows that $\lambda$ must be nonnegative, and then working as in Robert's answer we get the desired inequality.
A: I'm going to use subscript notation for partial derivatives, since that makes the Latex input easier on my 'droid, from which I write this response.  Thus I'll use
$f_x = \frac{\partial f}{\partial x}$,
$f_{xy}= \frac{\partial f}{{\partial x}{\partial y}}$,
and so forth.
These things being said, at even a local minimum of $f$ we must have 
$\nabla f = (f_x, f_y)   = 0$.
Now, since I don't want to go into the full, gory details and horrendous one-finger 'droid typing a  full $\epsilon$-$\delta$ proof would require, I'm going to beg your indulgence and ask you to try and see intuitively that, when $\nabla f(x_0, y_0) = 0$, most of the action, in terms of $f$ increasing or decreasing, is governed by the matrix of second derivatives
$H = \begin{bmatrix} f_{xx} & f_{xy} \\
f_{yx} & f_{yy} \end{bmatrix}$.
In particular, if we look at the eigenvalues and eigenvectors of $H$ at the point $ (x_0, y_0)$, we see (intuitively) that if
$Hv_1 = \lambda v_1$
with $\lambda > 0$, then moving in the direction of $v_1$ from $(x_0, y_0)$ will cause $f$ to increase, at least for a little while, at least near $(x_0, y_0)$.  Likewise, if $\lambda < 0$, $f$ will decrease.  This stuff has all been written up in great detail in the 
literature; you might have a look at John Milnor's little book Morse Theory for a really solid treatment.  The point is we must have $\lambda \ge 0$ for each eigenvalue of $H$ if we are truly at a local minimum.  So let's look at the characteristic equation of $H$, which the eigenvalues satisfy.  It is
$\det (H - \lambda I) = \lambda^2 - (f_{xx} + f_{yy}) \lambda + f_{xx}f_{yy} - f_{xy}^2 = 0$;
the quadratic formula yields
$\lambda = \frac{1}{2}((f_{xx} + f_{yy}) \pm  \sqrt{(f{xx} + f_yy)^2 - 4(f_{xx}f_{yy} - f_{xy}^2)}$,
and to ensure this is non-negative we need,
since the square root term is always non-negative, 
$f_{xx} + f_{yy} \ge \sqrt{(f_{xx} + f_{yy})^2 - 4(f_{xx}f_{yy} - f_{xy}^2)}, \tag{Z}$
and now its just algebra!  Square and regroup a few terms and you'll find, sure enough, that (Z) becomes
$f_{xx}f_{yy} \ge f_{xy}^2$.
QED!
