r^d \left(\frac{s}{s-r} + \frac{3 \rho}{rm}\right)^d \leq \nu m^d r^d \left(1 + \frac{3r +
6 d \rho}{s-r}\right)$$ where the final inequality follows because $(1 +x)^d \leq 1 + (2d)x$ for all sufficiently small $x$, and the fact that $rm \geq s-r$ by definition of $m$. Substituting this into the definition of $\tilde{\mu}_s$ produces the desired inequality on $\tilde{\mu}_s$ (assuming, without loss, that $r\le \rho$).

To show the lower bound on $\tilde{\tau}_s$, let $B$ be a ball of radius $r/2$. $\mathfrak{U}(B)$ will have diameter at most $s$, by definition of the metric, and its measure will be at least $\tau$. Converting this to cardinality gives the desired lower bound. For the upper bound, let $\mathfrak{W}$ be a set of indices such that $$\lambda \left(\mathfrak{U}(\mathfrak{W})\right) \geq \tau \left(1 + \frac{ C}{s} \right)$$ Applying the isodiametric inequality for finite dimensional normed spaces [@Burago p. 93] and choosing $C$ and $s_0$ sufficiently large gives $$\text{diam}(\mathfrak{U}(\mathfrak{W})) \geq r \left(1 + \frac{ C}{s} \right)^{1/d} \geq r + \frac{4 \rho}{m}$$ This implies that the diameter of $\mathfrak{W}$ is at least $s+1$ in the $d$ metric, and therefore the set cannot be a maximal clique set. Translating the measure bound to a cardinality bound on the size $\mathfrak{W}$ gives 