Minimum ratio between surface and volume in a riemannian manifold In an euclidean three - dimensional space the sphere is the geometric figure with the minimum ratio $R=\frac{S}{V}$ with $S=4\pi r^2$ and $V=\frac{4}{3}\pi r^3$, so we have: $$R=\frac{1}{3}r$$ where $r$ is the radius of the sphere. My question is: given a generic metric tensor $g_{\mu\nu}$ for which $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ how is it possible to find the surface with minimum $R$? Thanks for any suggestion

Among surfaces of revolution, it is not only the sphere but all DeLaunay unduloids share minimum surface area for given volume or, S/V property. It comes out as solution for an iso-perimetric ( calculus of variations) problem, meridians have a constant mean curvature property.

That is, k1 + k2 = constant is a a more general DeLaunay result than k1 = k2 = constant, the result for a sphere. ( k1,k2 are principal curvatures).