$\DeclareMathOperator\diam{diam}\DeclareMathOperator\inrad{inrad}$There is a isoperimetric inequality (conjectured by Sikorav and proven by Zuk [Topology 39 (2000) 947–956 ]) which holds in every Cayley graph of a group (with modest assumptions, such as countability of the group and finiteness of the generating set $S$), namely: $$ |A| \leq \diam(A) |\partial A| $$ Here $\partial A$ is the boundary of the set $A$ (the vertices which are adjacent to an element of $A$ but are not in $A$).

Erschler [Geometriae Dedicata 100: 157–171, 2003] proved that there are even groups where a stronger inequality holds: $$ |A| \leq K \; \diam(A)^k |\partial A| $$ where $K>0$ and $k\in {]0,1[}$.

My question regards a different strengthening of this inequality: given a [Cayley graph of a] group and a set $A$ whose complement has no finite connected components, are there $K,k >0$ so that $$ (?) \qquad |A| \leq K \; \inrad(A)^k |\partial A| ? $$ where $\inrad$ denotes the inradius, i.e. $\inrad(A) = \max \lbrace r \in \mathbb{N} \mid \exists a \in A$ so that $ B_r(a) \subset A \rbrace$ ; $B_r(a)$ being the ball of radius $r$ around $a$.

Remarks:

• $(?)$ holds for any non-amenable group, since, for such groups there is a $K$ so that $|A| \leq K |\partial A|$.

• $(?)$ holds for any group of polynomial growth (choose $K$ and $k$ so that $|B_{r+1}(a)| \leq Kr^k$ then $A$ may be covered by balls of radius $\inrad(A)+1$ centered on the vertices of $\partial A$.

• The restriction that the complement of $A$ has no finite connected components might not be necessary. As far as I know $k=1$ might also always work.

• If the inequality $(?)$ for two groups, the it holds for their wreath product. In particular, the inequality holds in the lamplighter group (with $k=1$).

• As far as I could check, the methods of Zuk and Erschler do not seem to yield this inequality.