# Prove the inequality for mean curvature of a closed surface $M$

Let $M$ be the closed surface generated by carrying a small circle with radius $r$ around a closed curve $C$ embedded in $\mathbb R^3$ such that the center moves along $C$ and the circle is in the normal plane to $C$ at each point. Prove that $$\int_M H^2 d\sigma \geq 2\pi^2,$$ and the equality holds if and only if $C$ is a circle with radius $\sqrt{2}r$. Here $H$ is the mean curvature of $M$ and $d\sigma$ is the area element of $M$.

Assume the curve $C$ is $\alpha(s):[a,b]\to \mathbb R^3$, so the surface can be represented by $$X(s,\theta)=\alpha(s)+r\cos(\theta)n(s)+r\sin(\theta)b(s).$$

I calculated $H$ but it's hard to figure out the integral :(

-
Not necessary to do the integral, only to bound it from below. This relates to Fenchel's theorem for the closed curve $C.$ –  Will Jagy Jun 21 '12 at 16:18
Note that the relationship to Fenchel's uses $\kappa \neq 0.$ –  Will Jagy Jun 21 '12 at 16:52

This is a classical result that you can find on p.274 of T.J. Willmore's Riemannian Geometry with a fairly complete calculation. The integral turns out to be quite tricky so even Willmore refers to Mathematica!

-
pages 113-114 in his Total Curvature in Riemannian Geometry. Very little calculation, he relates it to Fenchel's theorem, pages 90 and 97. –  Will Jagy Jun 21 '12 at 15:48