Let $P=\{x\in \mathbb{R}^n\mid a_i^\top x\leq b_i,\, i=1,\ldots,m\}$ be a bounded convex polytope, $\|a_i\|_2=1,\, i=1,\ldots,m$. Let us define the width of $P$ in the direction $a_i$ as $$ \operatorname{width}(a_i):= \max_{x\in P}a_i^\top x - \min_{x\in P}a_i^\top x, $$ and the smallest width $w:=\min_{1\leq i\leq m}\operatorname{width}(a_i)$. Is it true that volume of $P$ is no less than the volume of a ball of radius equal to half of the smallest width, i.e., $\operatorname{vol}P \geq \operatorname{vol}B_r$ with $r:=\frac{w}{2}$? If not, are there any similar relations? Can you recommend a book that could help answer the question?
Update: I checked the condition in $\mathbb{R}^2$ for regular $m$-polygons and it holds with $r:=\frac{w}{2}$ for even $m$, but not for odd $m$. Taking, for example, $r=\frac{w}{3}$ we can make it applicable for odd $m$ as well.
- Even $m$: inscribed circle touches bottom and top sides, thus $w$ equals its diameter, $w=\cot \frac{\pi}{m}$ (assume side length is 1). Area of the polygon is $\operatorname{vol}P=\frac{m}{4}\cot \frac{\pi}{m}$. Therefore, $$ \frac{\operatorname{vol}B_{w/2}}{\operatorname{vol}P}=\frac{\frac{\pi}{4} \cot^2\frac{\pi}{m}}{\frac{m}{4}\cot \frac{\pi}{m}}=\frac{\pi}{m}\cot \frac{\pi}{m} <1 $$ since $\cot t < \frac{1}{t}$.
- Odd $m$: width of the polygon equals $w=\frac{1}{2} \csc\frac{\pi}{m}\cdot (1+\cos \frac{\pi}{m})$, therefore, $$\frac{\operatorname{vol}B_{w/3}}{\operatorname{vol}P}=\frac{\frac{\pi}{9}\frac{1}{4} \csc^2\frac{\pi}{m}\cdot (1+\cos \frac{\pi}{m})^2}{\frac{m}{4}\cot \frac{\pi}{m}}=\frac{1}{9} \frac{\pi}{m}\frac{(1+\cos \frac{\pi}{m})^2}{\sin \frac{\pi}{m} \cos \frac{\pi}{m}} \leq \frac{1}{9} \frac{(1+\cos \frac{\pi}{m})^2}{\cos^2 \frac{\pi}{m}},$$ where the inequality is due to $\sin t \geq t \cos t$. Note that $m\geq 3 \Rightarrow \cos \frac{\pi}{m} \in [\frac{1}{2},1)$. The function $f(y):=\left(\frac{1+y}{y}\right)^2$ decreases on $[\frac{1}{2},1)$, therefore, $$ \frac{\operatorname{vol}B_{w/3}}{\operatorname{vol}P} \leq \frac{1}{9} \left(\frac{1+1/2}{1/2}\right)^2 = 1. $$
My intuition is that irregular polygon should have a larger volume than the regular one, provided the smallest width is the same. Moreover, I hope this can be extended to higher dimensions.
Update 2: I was initially interested in relation between polytope volume and width as a tool in establishing a result which is out of this post's scope. I recently found another way (unrelated to this relation) to obtain that result and I am therefore not working on this question anymore.