a diameter-perimeter-area inequality for convex figures - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:35:23Z http://mathoverflow.net/feeds/question/120529 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120529/a-diameter-perimeter-area-inequality-for-convex-figures a diameter-perimeter-area inequality for convex figures filipm 2013-02-01T16:34:59Z 2013-02-02T01:52:33Z <p>Is the following inequality known? I believe it's true, but I could find no reference. </p> <blockquote> <p>For any convex body $C$ in the plane we have $$\left(4-\frac{8}{\pi}\right)area(C)\leq diam(C)(per(C)-2diam(C)).$$</p> </blockquote> <p>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. </p> <p>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 <a href="http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1929_3_46_/ASENS_1929_3_46__345_0/ASENS_1929_3_46__345_0.pdf" rel="nofollow">here</a>. </p> http://mathoverflow.net/questions/120529/a-diameter-perimeter-area-inequality-for-convex-figures/120563#120563 Answer by Dmitri for a diameter-perimeter-area inequality for convex figures Dmitri 2013-02-01T23:59:30Z 2013-02-02T00:44:42Z <p>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 $4\varepsilon^2$.</p> http://mathoverflow.net/questions/120529/a-diameter-perimeter-area-inequality-for-convex-figures/120566#120566 Answer by Connor Mooney for a diameter-perimeter-area inequality for convex figures Connor Mooney 2013-02-02T00:39:36Z 2013-02-02T00:39:36Z <p>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 $\epsilon^2$.</p> <p>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.</p>