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 $${\rm mean}({\rm vol}(A+B)^{1/d},{\rm vol}(A\oplus B)^{1/d})\le{\rm vol}(A)^{1/d}+{\rm vol}(B)^{1/d}.$$

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    $\begingroup$ I don't know anything about the "sum" you define, but Erwin Lutwak introduced something similar, called the radial sum, and it is one of the basic ingredients of what he called "dual Brunn-Minkowski theory". If you google those terms, you can find lots of papers and books that discuss this. $\endgroup$ – Deane Yang Jan 3 '17 at 20:35
  • $\begingroup$ I think it is the radial sum, is it not? $\endgroup$ – alvarezpaiva Jan 12 '17 at 12:57

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

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

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