Reading Difficulty: 5/5
Prerequisites: Understand literally anything in the first paragraph

If you have studied PDEs, you may be familiar with the Laplace problem with a Dirichlet boundary condition:
The boundary condition here describes pointwise values that needs to take, and is hence called a Dirichlet condition. There are other sorts of boundary conditions, however, and one such sort is a von Neumann boundary condition, which imposes a condition on the gradient of on the boundary:
where denotes the unit outward normal to the relevant boundary (which, here, is ).

Let us take to be sufficiently nice. Then an obvious necessary condition for a solution to exist is found via the Divergence theorem. To be precise, if solves the above problem then we must have
where denotes the -dimensional Hausdorff measure. For the record, we will be using to denote the -dimensional Lebesgue measure.

It turns out that the condition is not only necessary but sufficient. (See https://math.stackexchange.com/questions/3214541/necessary-and-sufficient-conditions-for-the-solution-of-neumann-problem) What is surprising is that this fact implies the Isoperimetric Inequality! Let be a bounded domain with smooth. We will prove that
for a constant depending only on , with equality exactly when is a ball. Equivalently, we will show that
where denotes the unit -ball.

Begin by finding the unique solution to the von Neumann problem
The constant was chosen carefully to ensure that we satisfy the necessary and sufficient condition for existence of the solution .

Now let be the set of points for which the tangent plane to the graph of at lies entirely below . That is, . What's important about is that is convex inside . This will be useful for taming its Hessian matrix, as we'll see soon.

Key Claim:

Proof. Here is the intuitive argument. Take and take a plane with gradient . Raise this plane from below the graph of until it hits the graph at a point . Since , and on we have from the von Neumann condition, it cannot be the case that . So . Clearly and , so .

The rigorous details go something like this.

As you might guess, this claim is precisely how we sneak in the ball into our equations. By using as a change of variables (!), we have that
where we write because may not be injective.

Now we must reason about the determinant of the Hessian . But recall that is convex in ! So in , must be positive semi-definite, so all its eigenvalues are non-negative everywhere. It follows that
by AM-GM. This gives
which rearranges to
Taking the th root concludes the proof.

This methodology is known as the ABP estimate, and the above application of this technique for proving the isoperimetric problem was discovered relatively recently by X. Cabre.