Integral of the distance function to the boundary of a planar set I have been stuck for a few days in a seemingly harmless question.
Given a simply connected open set $\Sigma\subset\mathbb{R}^2$, with smooth boundary $\partial\Sigma$, I am interested in estimating
$$
\int_{\Sigma}d(x,\partial\Sigma)\;dx,
$$
where $d(x,\partial\Sigma)=\inf_{y\in\partial\Sigma}|x-y|$.
I would love to be able to prove that
$$
\int_{\Sigma}d(x,\partial\Sigma)\;dx\ge\frac{2}{3} \frac{|\Sigma|^2}{|\partial\Sigma|}.
$$
Equality holds for $\Sigma$ being a disk and violating the smoothness assumption for $\Sigma$ being a square.
The only thing that seems to be doing on the right hand side of the inequality is using the Isoperimetric Inequality to say
$$
\frac{2}{3} \frac{|\Sigma|^2}{|\partial\Sigma|}\le \frac{1}{3\sqrt{\pi}}|\Sigma|^{\frac{3}{2}},
$$
with equality, if and only if, $\Sigma$ is a disk.
Ideas or counter-examples are very much welcomed.
 A: The correct constant is $1/2$. Let $L(s)$ be the perimeter of the set of points whose distance to the boundary is $\ge s$. For simply-connected domains $L(s)$ is non-increasing, the integral of the distance to the boundary is $\int_0^\infty sL(s)\,ds$ and the area is $\int_0^\infty L(s)\,ds$. Now we want to minimize the first integral given the second one and $L(0)=\max_s L(s)$. Clearly, the worst case is when we push all the mass to the left as much as we can, so $L(s)=L(0)$ up to some point and then $0$. That gives you $1/2$ for the ratio in question and that scenario is almost realized by a long narrow rectangle, as Nate pointed out.
A: Mark's example is essentially two rectangles.  But the inequality is stable under doubling the set, so if it fails for two rectangles, it ought to also fail for one.
And it does.  Consider the rectangle $\Sigma = [0,1] \times [0,N]$ where $N$ is large.  On $\Sigma' = [0,1] \times [1, N-1]$, we are always closer to the sides of the rectangle than to the top or bottom, so the average distance to the boundary is $\frac{1}{4}$, and we have $$\int_{\Sigma'} d(x, \partial \Sigma)\,dx = \frac{1}{4}(N-2) \approx \frac{N}{4}.$$  The contribution from $\Sigma \setminus \Sigma'$ is at most $2$ which is negligible, so the left side is approximately $\frac{N}{4}$.  Now $|\Sigma| = N$ and $|\partial \Sigma| = 2N+2$.  So the right side is $\frac{2N^2}{3(2N+2)} \approx \frac{N}{3}$ and the inequality will fail for sufficiently large $N$.  You may then round off the corners for a smooth convex counterexample.
We could still ask whether it might hold with a different constant in place of $\frac{2}{3}$.  Based on the examples so far, $\frac{1}{2}$ is still plausible.
A: A simply-connected counterexample: Consider a unit square; round the corners by replacing them with quarter-circles of radius some tiny $\epsilon$.  Now deform the top edge by cutting out the $4\epsilon$ segment centered at the top of the top edge, and replacing it with a line from the left of the cut, perpendicular to the top, extending to within $2\epsilon$ of the bottom side, plus a line of length $4\epsilon$ parallel to the top moving to the right, plus a line connecting the end of that segment with the right end of the cut-out segment on the top. Finally, use $\epsilon$ quarter-circle replacements to round all the right angles you have introduced.  This gives an open simply-connected area  with a smooth boundary.
For very small $\epsilon$, the area looks like a square divided into two rectangles by a double-line edge douwn the middle.  Then:


*

*The integral of the distance breaks up into 16 right isoceles triangles of side $\frac14$, each of which contributes $\frac{1}{396}$ plus four rectangles of sides $\frac14$ and $\frac12$, each of which contributes $\frac{1}{32}$.  The total is
$$
\int_\Sigma d(x,\partial \Sigma) dx =\frac5{48}
$$

*The area $|\Sigma| = 1 - O(\epsilon)$

*The boundary length is  $|\partial\Sigma| = 6 - O(\epsilon)$
This is a counterexample because 
$$
\frac23 \frac{|\Sigma|^2}{|\partial\Sigma|} = \frac23 \cdot  \frac{1^2}{6} + O(\epsilon) = \frac19 + O(\epsilon) > \frac5{48}
$$
I'm not sure, but your inequality might hold for convex open sets (which have smooth boundaries).
A: Consider an $s\times2$ rectangle with semicircles of radius 1 on the ends.  The inequality is false for such shapes.
In the rectangle, the average distance to the boundary is 1/2.
In the semicircles, the average distance to the boundary is 1/3.
So the inequality is
$$\frac{2s}{2} + \frac{\pi}{3} \ge  
 \frac{2}{3} \frac{(2s+\pi)^2}{2s+2\pi},$$
which is easily seen to be false for large $s$. 
