$\DeclareMathOperator\diam{diam}\DeclareMathOperator\inrad{inrad}$There is a isoperimetric inequality (conjectured by Sikorav and proven by Żuk (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 $\partial A$. (This most often gives a bound with $k>1$, and is probably never sharp.)
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 $(?)$ holds for two groups, then 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 Żuk and Erschler do not seem to yield this inequality. In short, it boils down to the fact, that, though the average of some data is a useful lower bound for the maximum, it is not a lower bound for the minimum (in these papers, the average is in terms of the distance to $|\partial A|$).
one can reformulate this inequality as trying to give a lower bound on $\inrad$ in terms of $\frac{|A|}{|\partial A|}$ (which is actually the angle I am interested in).
Let the marrow $M(A) = \lbrace a \in A \mid d(a,\partial A) = \inrad(A) \rbrace$ be the set of elements where the inradius is attained. The following inequality seems to be related: is there a $K>0$ so that $$ (??) \qquad |M(A)| \leq K |\partial A| ? $$ though that may be a completely different question.
there are [infinite] regular graphs where both $(?)$ and $(??)$ fail. An example is the following: start with an tree which is 3-regular but has it leaves on $\mathbb{Z}$ and only one point at infinity (i.e. take $\mathbb{Z}$ as the leaves; these leaves are only connected to another copy of $\mathbb{Z}$ in the following way: $n$ is connected to $\lceil n/2 \rceil$; repeat ad infinitum). Take 3 copies of this tree and join them at their leaves. The resulting graph has an abysmally bad isoperimetric ratio (it is possible to disconnect arbitrarily large sets with 3 vertices). As such its not too surprising that $(?)$ and $(??)$ fail.