Around Brunn-Minkowski inequality

Question

Let me recall the Brunn-Minkowski inequality, which states concavity of ${\rm vol}^{1/d}$ for domains in ${\mathbb R}^d$: $${\rm vol}(A+B)^{1/d}\ge{\rm vol}(A)^{1/d}+{\rm vol}(B)^{1/d},$$ with equality only if $A$ and $B$ are homothetic (that is $B=\lambda A+v$).

Suppose now that $A$ is star-shaped about the origin. It can be described from a function $f:S^{d-1}\rightarrow(0,+\infty)$ by $$A=A^f:=\{r\omega\,|\,\omega\in S^{d-1},\,0\le r<f(\omega)\}.$$ If $A^g$ is star-shaped too, we can define another "sum" by $A^f\oplus A^g:=A^{f+g}$. On the one hand, we have $A\oplus B\subset A+B$ for star-shaped domains. On the other hand, the formula ${\rm vol}(A^f)^{1/d}=c_d\|f\|_{L^d}$ and the Minkowski inequality give $${\rm vol}(A\oplus B)^{1/d}\le{\rm vol}(A)^{1/d}+{\rm vol}(B)^{1/d},$$ with equality only if $A$ and $B$ are homothetic (here $B=\lambda A$).

Question: How does ${\rm vol}(A)^{1/d}+{\rm vol}(B)^{1/d}$ compare with means (say arithmetic or geometric) of ${\rm vol}(A+B)^{1/d}$ and ${\rm vol}(A\oplus B)^{1/d}$ ?

The case with $B=A+v$ suggests that if an inequality holds true, it must be of the form $${\rm mean}({\rm vol}(A+B)^{1/d},{\rm vol}(A\oplus B)^{1/d})\le{\rm vol}(A)^{1/d}+{\rm vol}(B)^{1/d}.$$

Answer 1

There is a easy counter example for both athematic mean and geometric mean.

Take $A=[0,1]\times [0,\delta]$, $B=[0,\delta]\times [0,1]$, $\delta <<1$, then $A+B=[0,1+\delta]\times [1+\delta,0]$, $A\oplus B$ shape is a cross, $\mu(A+B)\sim 1+O(\delta)$, $\mu(A\oplus B)=2\delta+O(\delta^2)$.

So $\mu^{\frac{1}{2}}(A\oplus B)=\sqrt{2\delta}+O(\delta)$, $\mu^{\frac{1}{2}}(A+B)\sim 1+O(\delta^{\frac{1}{2}})$.

So, $\mu^{\frac{1}{2}}(A\oplus B)+\mu^{\frac{1}{2}}(A+B)\sim 1+O(\delta^{\frac{1}{2}})>>\mu^{\frac{1}{2}}(A)+\mu^{\frac{1}{2}}(B)$.

$\sqrt{\mu^{\frac{1}{2}}(A\oplus B)\cdot \mu^{\frac{1}{2}}(A+B)}\sim O(\delta^{\frac{1}{4}})>>\mu^{\frac{1}{2}}(A)+\mu^{\frac{1}{2}}(B)$