# a diameter-perimeter-area inequality for convex figures

Is the following inequality known? I believe it's true, but I could find no reference.

For any convex body $C$ in the plane we have $$\left(4-\frac{8}{\pi}\right)area(C)\leq > diam(C)(per(C)-2diam(C)).$$

If true, this would be tight, with equality when $C$ is a disk. If it turns out not to be true, then it still makes sense to look for the best constant with the $area(C)$ term.

The following similar inequality $diam(C)(per(C)-2diam(C))\leq\frac{4}{\sqrt3}area(C)$ (equality when $C$ is an equilateral triangle) is definitely known and is proven here.

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The paper, "Inequalities for Convex Sets," by Scott and Awyong, Journal of Inequalities in Pure and Applied Mathematics, might help, although I don't see your inequality there (but maybe I missed it amidst the welter of notation). PDF download: emis.de/journals/JIPAM/images/016_99_JIPAM/016_99.pdf –  Joseph O'Rourke Feb 1 '13 at 16:53
You should also check the book "Geometric Inequalities" by Burago and Zalgaller. –  Misha Feb 1 '13 at 18:16
I had checked before both references (by Scott and Awyong and by Burago and Zalgaller), but I couldn't find what I need. –  filipm Feb 1 '13 at 19:04

This inequality is not true. Consider the rectangle on $\mathbb R^2$ with vertices $(\pm 1, 0)$, $(0, \pm \varepsilon)$. Then on the left you have $2\varepsilon(4-8/\pi)$ on the right you have approximatively $4\varepsilon^2$.

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Your numbers don't look right to me, but this does look like a counterexample. –  Deane Yang Feb 2 '13 at 0:06
On second thought, this does not look like a counterexample. The left side appears to be equal to $4\epsilon(4−8/\pi)\simeq5.8\epsilon$ and the right side appears to be equal to $2\sqrt{1+\epsilon^2}(4(1+\epsilon)-4\sqrt{1+\epsilon^2}) \simeq 8\epsilon$. –  Deane Yang Feb 2 '13 at 0:20
Sorry, there was a misprint in my answer. now it is correct –  Dmitri Feb 2 '13 at 0:45
I also called rectangle what should be called a rombus :) –  Dmitri Feb 2 '13 at 0:52
Yes. I misinterpreted your answer. –  Deane Yang Feb 2 '13 at 1:53

Here's a variation on Dmitri's idea that works as a counterexample: Take the rhombus with long diagonal $2$ and short diagonal $2\epsilon$. Then the area (LHS) grows like $\epsilon$, but the perimeter minus twice diameter goes like $$4\sqrt{1+\epsilon^2}-4$$ which grows like $\epsilon^2$.

The idea is that by moving out $\epsilon$ in the "center" rather than the edges will give quadratic growth of the perimeter rather than linear.

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Connor, that is of course correct. In fact, I was thinking of exactly this example but for some reason (I guess to make the answer as short :) ) as possible put the vertices of the rombus in $\pm 1, \pm varepsilon$) instead of what I had in mind. –  Dmitri Feb 2 '13 at 0:55
Yes, in restrospect I should have posted this as a comment on your answer- cheers! –  Connor Mooney Feb 2 '13 at 1:16