Let $A$ be a closed subset of $\mathbb{R}^n$. For $r>0$, let $A_r$ be the $r$-neighborhood of $A$, namely the set $\{x:\operatorname{dist}(x,A)\le r\}$. Is the function $f(r) = \mu(A_r)^{1/n}$ concave? ($\mu$ is the Lebesgue measure.)
Context
This came up in the discussion of George Lowther's answer, where I asserted that the concavity of $f$ follows from the Brunn-Minkowski inequality. As George Lowther pointed out, this approach seems to require $A$ to be convex. Here's a proof for convex $A$.
Since $A$ is convex, we have $A=\frac12(A+A)$, using Minkowski/vector addition. Let $B$ be the closed unit ball (also convex). For $r,s>0$ we have $$ A+\frac{r+s}{2}B = \frac12 (A+A)+\frac r2 B + \frac s2 B = \frac12(A+rB)+\frac12(A+sB) $$ Taking the $n$th root of volume on both sides and using the Brunn-Minkowski inequality, we obtain $$ f((r+s)/2) = \mu\left(\frac12(A+rB)+\frac12(A+sB)\right)^{1/n} \ge \frac12 \mu(A+rB)^{1/n}+\frac12 \mu(A+sB)^{1/n} $$ proving that $f$ is midpoint-concave. Since it's also monotone, it is concave.
The proof falls apart when $A$ is not convex, since the inclusion $A\subset \frac12(A+A)$ goes the wrong way. But I don't see a counterexample.
No, $f$ does not have to be concave. For a counterexample in dimension $n=2$, let $A$ be the union of the closed unit disc centred at the origin and a single point $P$ with $\lVert P\rVert > 1$. We can compute the area of $A_r$ easily for small $r$ (specifically, $2r\le\lVert P\rVert-1$), $$ \mu(A_r)=\pi(1+r)^2+\pi r^2=\pi(1+2r+2r^2). $$ Differentiating $f(r)=\sqrt{\pi(1+2r+2r^2)}$ twice, $$ f^{\prime\prime}(r)=\frac{\sqrt\pi}{(1+2r+2r^2)^{3/2}} > 0. $$ So, $f$ is strictly convex for $r\le(\lVert P\rVert-1)/2$..