Minimum ratio between surface and volume in a riemannian manifold

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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