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If $r\ge \sqrt{2}$ and a convex polytope satisfies $r^{-1}{B}^n\subseteq P\subseteq B^n$, then it has at least $(1-r^{-2}{)}^{-n/2}\ge e^{n/(2r^2)}$ facets. Its outward facet normals define spherical caps covering the unit sphere; the spherical cap area upper bound supplies the estimate.

Let $m$ be the number of facets of the convex polytope $P$. Its facet description can be writtenBecause $r^{-1}{B}^n\subset P$, each supporting closed half-space has $h_j\ge 1/r$. For any $\theta \in S^{n-1}$, follow the ray $t\theta$ to its last point in $P$. Its radius $t_{\theta }$ is at most one because $P\subset B^n$. At least one facet is active there, givingThus the unit sphere is covered by the $m$ spherical caps centred at $u_j$ with angular radius $\rho =\arccos (1/r)$. Since $r\ge \sqrt{2}$, this radius lies in $[\pi /4,\pi /2)$ and the spherical cap area upper bound applies. Subadditivity of surface area givesUsing $\log (1-x)\le -x$ for $0<x<1$,The facet lower bound for ball approximations is exponential in the dimension.

Use the same normalized spherical cap area $q$, constant $c_n$ and function $F={\sin }^n\rho -q$ as above. Put ${\rho }_0=\arcsin (1/\sqrt{n})$. The supplied ratio estimate givesThe middle inequality holds for every $n\ge 2$, since $n+1\le 2\pi (n-1)$. Applying the upper projection bound for the spherical cap at ${\rho }_0$ yieldsso $F({\rho }_0)\ge 0$.

RecallAt ${\rho }_0$ the expression in parentheses is positive, because $n\sin {\rho }_0\cos {\rho }_0=\sqrt{n-1}$ and $(n-1)c_n<\sqrt{n-1}$. It increases up to $\pi /4$ and then decreases strictly to a negative value at $\pi /2$. Hence it crosses zero exactly once on $[{\rho }_0,\pi /2]$, from positive to negative. The minimum of $F$ is again an endpoint value; $F({\rho }_0)\ge 0$ and $F(\pi /2)=1/2$. This proves the stronger range of the spherical cap area upper bound:

Normalize the surface area of the spherical cap by writingTwo elementary descriptions of a spherical cap are useful. The spherical polar coordinates formula givesProjecting orthogonally onto the last $n-1$ coordinates givesIndeed, the upper hemisphere is the graph $x_1=\sqrt{1-\Vert x^{\prime }{\Vert }^2}$, whose surface area factor is $(1-\Vert x^{\prime }{\Vert }^2{)}^{-1/2}\le 1/\cos \rho$ on $\Vert x^{\prime }\Vert \le \sin \rho$. This establishes the upper projection bound directly.

Set $F(\rho )={\sin }^n\rho -q(\rho )$. The supplied ratio bound implies $c_n<\sqrt{(n+1)/(2\pi )}/n\le 1/2$ for $n\ge 2$. Consequently, at $\rho =\pi /4$,Also,On $[\pi /4,\pi /2]$, the expression in parentheses is strictly decreasing; it starts positive and ends negative. Thus $F$ first increases and then decreases, so its minimum on this interval is at an endpoint. Since $F(\pi /2)=1/2$, both endpoint values are nonnegative. The spherical cap area upper bound follows: