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 :(