# Another diameter-perimeter-area inequality

Recently I learnt that $$\DeclareMathOperator{\diam}{diam}\DeclareMathOperator{\per}{per}\DeclareMathOperator{\area}{area} \inf\frac{\diam(C)(\per(C)-2\diam(C))}{\area(C)}=0$$ where the infimum is taken over all plane convex bodies $$C$$ (say, with non-zero area). In other words, there is no non-trivial (with $$c>0$$) inequality of the form $$\diam(C)(\per(C)-2\diam(C))\geq c\cdot \area(C),$$ that would hold for all $$C$$.

Now I'm wondering about inequalities of the form $$\diam(C)(\per(C)-(2-\epsilon)\diam(C))\geq c_{\epsilon}\cdot \area(C)$$ for $$\epsilon>0$$. Clearly, such an inequality is true for $$c_\epsilon=\frac{4\epsilon}{\pi}$$ (since $$\per(C)\geq2\diam(C)$$ and $$\diam(C)^2\geq4/\pi\cdot \area(C)$$).

The question is: what is the best $$c_{\epsilon}$$ for which the inequality holds? When would equality be achieved (it might depend on $$\epsilon$$)?

## 1 Answer

Let us call a circular segment a thing like this: (| : https://en.wikipedia.org/wiki/Circular_segment

I think that the best possible constant will be achieved on doubled circular segments, i.e. things like this : ()

Indeed suppose you fix the area and diameter of you domain, and try to minimise its perimeter. Then the diameter cuts your domain into two halves. If you fix the area of one half (and minimise its perimiter), then this is a classical fact that half will be a circular segment (|.

So one juts needs to take all the formulas from wiki (on the area and perimeter of circular segments) and find the minimum.