In-radius of a convex set with prescribed volume

Question

Assume you have an $n$ - dimensional convex body $C$ with prescribed volume $V$. Let $diam(C)$ be the diameter of $C$. Is there an inequality relating the in-radius of $C$ with these quantities?

Answer 1

The inradius can be arbitrarilty small by taking a long but thin cuboid.

The largest inradius is achieved with a ball whose volume is $V$. Say it is has radius $r_V$. Any other convex set $C$ with the same inradius then contains a ball $B$ of radius $r_V$. But then $volume(C\backslash B)=0$. I claim $C\backslash B$ is contained inside the closure of $B$. Suppose not, then there is some point $c\in C\backslash B$ such that the distance from $c$ to the center of $B$ is greater than $r_V$. But then as $C$ is convex, $C$ contains a cone extending from $c$ to the surface of $B$. This cone has positive volume and is in $C\backslash B$, contradiction. Thus, any such $C$ is contained in the closure of $B$.

Overall, we have shown that the ball achieves the inequality and any such convex set is contained in the closure of the inradius ball.

In short, $inradius(C)\leq r_V$. The formula for the radius in terms of volume can be found here: https://en.wikipedia.org/wiki/Volume_of_an_n-ball