Prove that volume of a ball in a polytope is very small An exercise in a book asks to prove that for a bounded convex polytope $P\subseteq\mathbb{R}^n$ defined as an intersection of $k$ closed halfspaces and for a unit ball $B^n$ contained in $P$ the following holds $$\frac{\lambda(B^n)}{\lambda(P)}\leq\left(\frac{c \ln k}{n}\right)^{n/2}$$ where $c$ is a suitable constant. It is also hinted to consider a suitable sphere $S$ such that the halfspaces complementary to those which define $P$ together cover half of $S$.
It will probably use the measure concentration theorem for the sphere: Let $A\subseteq S^{n-1}$ be a measurable set with $\mu(A)\geq\frac12$, let $A_t$ denote the $t$-neighbourhood of $A_t=\left\{x\in S^{n-1} : d(x,A)\leq t \right\}$. Then $1-\mu(A_t)\leq 2e^{-t^2n/2}$.
The spheric measure $\mu(X)$ for $X\subseteq S^{n-1}$ is defined as the fraction of the volume of the unit ball occupied by the points connecting $X$ to $0$, that is $$\mu(X)=\frac{\lambda\left(\{tx : x\in X, 0\leq t\leq1 \}\right)}{\lambda(B^n)}$$
Thank you for any assistance.
 A: I might have found the solution myself:
We have the polytope $P$ and the unit ball $B^{n}$, suppose that we construct a sphere $S$ with the same center as the unit ball and radius $r$ such that the halfspaces complementary to those which define $P$ together cover exactly half of $S$ (measured by $\mu$).
Now we transform the space so that $S$ becomes a unit sphere $S^{n-1}$. $B^{n}$ becomes a ball with radius $1/r$. Let's consider $C\subseteq S^{n-1}$ the covered part of the sphere. We know that each point $p\in C$ is cut of from $P$ by some hyperplane $h$. Also the hyperplane is in distance at least $1/r$ from the origin (because of the ball inside). From the MCT we may deduce that the hyperplane $h$ cuts of at most $2e^{-n/2r^{2}}$ of the sphere (for $A$ we use a hemisphere bounded by a hyperplane parallel to $h$ going through the origin). We need to cover/cut off the whole $C$ and $\mu\left(C\right)=1/2$, so we need at least $e^{n/2r^{2}}$ such hyperplanes, i.e. $k\geq e^{n/2r^{2}}$. We solve the inequality for $r$ and get $r\geq\left(\frac{n}{2\ln k}\right)^{1/2}$. From the construction of $S$ we know that at exactly half of the sphere (measured by $\mu$) is still inside $P$ and so also at least half of the volume of the ball of radius $r$ enclosed by $S$ is still in $P$. That means $\lambda\left(P\right)\geq\frac{1}{2}r^{n}\lambda\left(B^{n}\right)$ i.e. $$\frac{\lambda\left(B^{n}\right)}{\lambda\left(P\right)}\leq2r^{-n}\leq2\left(\frac{2\ln k}{n}\right)^{n/2}\leq4^{n/2}\left(\frac{2\ln k}{n}\right)^{n/2}=\left(\frac{c\ln k}{n}\right)^{n/2}.\;\square$$
