Let $f$ be convex, twice continuously differentible on $D=\{(x,y);x^2+y^2\leq 1\}$. Suppose that $f\geq 0$ on $\{(x,y);x^2+y^2=1\}$. Show that $$f(0)\geq -\frac{1}{\sqrt{\pi}}\left[\iint_{D} f_{xx}f_{yy}-f_{xy}^2\right]^{1/2}$$ Here $f_{xx}$ means $\partial^2f/\partial x\partial y$, and etc.
At first glance, let $g(t)=f(t\cos \theta,t\sin\theta)$, and $g(0)\geq g(1)+g'(1)(-1)$. What to do next? How to find the $1/2$ in the problem.
This is proved by a geometric argument which usually appears along with the names Alexandrov, Bakelman, and Pucci. The idea is to look at the gradient $\nabla u$ as a map. Let $\Omega$ be the image of $D$ under this map.
Let $M = -f(0)$. For every vector $\xi$ with $|\xi|<M$ the function $v(x) = f(x)-\xi\cdot x$ is convex in $D$ and satisfies $$ \min_{\partial D} v \ge -|\xi| > -M = v(0) $$ Therefore $v$ attains its minimum in $D$, which implies $\nabla f(x) = \xi$ for some $x$. We conclude that $\Omega$ contains an open disk of radius $M$. Thus, the area of $\Omega$ is at least $\pi M^2$.
On the other hand, this area can be computed by integrating the Jacobian of the gradient map, which is the Hessian $\det D^2 f$ (nonnegative by the convexity of $f$): $$ \pi M^2 \le |\Omega| = \iint_D \det D^2 f $$ which proves the claim.
Section 9.1 of Gilbarg-Trudinger has more general results.