a diameter-perimeter-area inequality for convex figures - MathOverflow

a diameter-perimeter-area inequality for convex figures

2013-02-01T16:34:59Z

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.

Answer by Dmitri for a diameter-perimeter-area inequality for convex figures

2013-02-01T23:59:30Z

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 $4\varepsilon^2$.

Answer by Connor Mooney for a diameter-perimeter-area inequality for convex figures

2013-02-02T00:39:36Z

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.