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If $\Sigma$ is a smooth embedded closed hypersurface in $\mathbb R^n$ with (normalized) mean curvature $H\le 1$ (the mean curvature of the unit sphere), then its ($(n-1)$-dimensional) area is at least the area of the unit sphere in $\mathbb R^n$. This is proved in Chavel's book "Isoperimetric inequalities" Theorem II.1.3.

The proof uses the idea that by restricting to the set of elliptic points (on which the Gauss map is surjective onto the sphere), the Gauss curvature $K$ is just the Jacobian determinant of the Guass map, and by the AM-GM inequality, it is less than $H^{n-1}$, which is less than $1$. The result follows by integration.

My question is, does this generalize to the hyperbolic space and the sphere as well? To be concrete, I suppose if $H\le \coth r$ for $\Sigma$ in the hyperbolic space, its area should be at least that of the sphere of radius $r$. Is it true? I don't see how the Gauss map in the proof above can be modified to give a proof of this assertion.

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    $\begingroup$ This bound was also shown (in Euclidean space) by Almgren in his paper "Optimal Isoperimetric Inequalities" (Almgren calls it the Area-mean curvature characterization of standard spheres). His argument actually works for any codimension and so should also give the result you asked about in spheres (by embedding the ambient sphere into euclidean space). I'm not sure what happens in hyperbolic space. $\endgroup$
    – RBega2
    Commented Oct 27, 2019 at 19:08

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