On a relation between volume of subsets of $\mathbb R^n$

Solution 1:

This is not true. In fact, you must have $p=q=r$ and $r \geqslant \frac 1 n$, see the Brunn-Minkowski theorem. (I feel like you must also have $r \leqslant \frac 1 n$ but I haven't though of a way to prove this). This all comes down to rescaling. Let $B_R$ denote a ball with radius $R>0$. I split the proof into several parts. For simplicity, I will also only deal with the case $p,q,r>0$ - with a suitable modification of the proofs below you should be able to obtain the same result.

Proof that $p=q=r$: It is an easy exercise to prove that $B_{R_1}+B_{R_2} = B_{R_1+R_2}$. Choose $R$ such that $\nu (B_R)=1$. Then for all $s,t>0$,\begin{align*} (s+t)^{nr}&= \nu (B_{(s+t)R})^r \\ &\geqslant \nu (B_{sR})^p + \nu (B_{tR})^q \\ &=s^{np} + t^{nq}.\tag{$\ast$} \end{align*} Sending $s \to 0^+$, in $(\ast)$ we obtain $$ t^{nr} \geqslant t^{nq} \qquad \text{ for all } t>0.$$ This is only true if $q=r$. Similarly, we may send $t\to 0^+$ in $(\ast)$ to obtain $p=r$.

Proof that $r \geqslant \frac 1 n$: Setting $s=\delta t$, $\delta>0$ in $(\ast)$ and, using that $p=q=r$, we obtain $$(1+\delta)^{nr} \geqslant \delta^{nr} +1 \qquad \text{ for all } \delta>0.$$ This is true if and only if $nr \geqslant 1$.