Let $F = \{P_1, P_2, \dots, P_n\}$ and let $\alpha_{ij}$ be the perpendicular plane bisector of $P_i$ and $P_j$ for $i \neq j$. Those planes define $n$ convex regions $R_1, R_2, \dots, R_n$, where $R_i$ is the intersection of the half-spaces determined by $\alpha_{ij}$ that contain $P_i$. Finally, let $A_r(i)$ be the intersection of $R_i$ with the sphere with center $P_i$ and radius $r$. Thus, since the (disjoint) union of the regions $R_i$ is the whole space, $V_r(F)$ is the disjoint union of $A_r(1), A_r(2), \dots, A_r(n)$.

Now, for each point $P_i$, apply a homothety with center $P_i$ and ratio $r/R < 1$. It is clear that the image of $A_R(i)$ is contained in $A_r(i)$, since a sphere with radius $R$ is taken to a sphere with radius $r$ and the planes $\alpha_{ij}$ are taken to planes $\alpha'_{ij}$ closer to $P_i$. Thus $\text{vol}(A_r(i)) \ge (\frac{r}{R})^3 \text{vol}(A_R(i))$ and the result follows by summing up these inequalities for $i = 1, 2, \dots, n$.