Concavity of the $n$th root of the volume of $r$-neighborhoods of a set
Solution 1:
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 $r\le(\lVert P\rVert-1)/2$..