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I recently saw this question, asking if the circle is the planar figure which has the least perimeter given the area. As far as I know, this is a classical problem, and the answer is affirmative. I am also fairly certain that the same is true for the ball in three dimensions: given a fixed volume, the solid with that volume and minimal area is the ball.

However, I realised with some surprise that I do not know if this generalises to higher dimensions. Hence the question:

Consider all $n$-dimensional manifolds with doundary $M \subset \mathbb{R}^n$ with $\operatorname{vol}(M) = 1$. Is it true that $\operatorname{vol}(\partial M)$ is minimised by the ball (with the appriate radius)?

I think this should be true, but I am also aware that high-dimensional geometry can produce very unexpected results, so I am far from being certain.

(Because everything happens in $\mathbb{R}^n$, I believe there is a natural notion of volume on submanifolds; I denote this volume by $\operatorname{vol}$)

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  • $\begingroup$ Do you really mean to say "volume to area"? In the body of the post the question is clearer: the volume of the manifold and the volume of its perimeter. $\endgroup$
    – rschwieb
    Commented May 12, 2013 at 12:02
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    $\begingroup$ See wikiepedia: Isoperimetric inequality $\endgroup$
    – Harald Hanche-Olsen
    Commented May 12, 2013 at 12:05
  • $\begingroup$ @rschwieb: Apologies for the vague wording of the topic, I hope it is better now. It would be more precise to say "$(n-1)$-dimensional volume of the boundary", I suppose, but hopefully it is understandable enough. $\endgroup$
    – Jakub Konieczny
    Commented May 12, 2013 at 12:07
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    $\begingroup$ For 3 dimension, it's indeed true, and it's the reason why soap bubbles and water drops are spherical! (In their steady state, at least). They minimize their surface area, since a larger area translates to a higher energy state in a way, and they do tend to be in the state with the lowest energy. $\endgroup$
    – fgp
    Commented May 12, 2013 at 12:23
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    $\begingroup$ This is the classical isoperimetric inequality. To get started, see the section on higher dimensions in the Wikipedia article, which has references to some textbooks where the proof can be found. $\endgroup$
    – Jim Belk
    Commented May 12, 2013 at 15:53

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