TITLE: Characterize the sphere using mean curvature.
QUESTION [5 upvotes]: We know the following result: if $\Sigma$ is a compact surface than
$$ \int_{\Sigma}H^2 \ge 4 \pi, $$
where with $H= \frac{1}{2}(\kappa_1+\kappa_2)$ we denote the main curvature. I have to prove that $\int_{\Sigma} H^2 = 4 \pi$ if and only if $\Sigma$ is a sphere $\mathbb{S}_R$. Now, if $\Sigma=\mathbb{S}_R $ then we know that $\kappa_1=\kappa_2=\frac{1}{R}$, so
$$ \int_{\mathbb{S}_R}H^2 d\mathcal{A}_{\mathbb{S}_R}=\frac{1}{R^2}  \int_{\mathbb{S}_R} 1 \,\,d\mathcal{A}_{\mathbb{S}_R} = \frac{4\pi R^2}{R^2}= 4 \pi .$$
How can I prove the other part of the proposition? Thank you!

REPLY [2 votes]: Recall that $K= k_1k_2$ and Gauss-Bonnet theorem implies $$ \int_S K
dS=2\pi \chi(S) $$
If $S$ is homeomorphic to $S^2$ then $$ \int_S K dS=4\pi $$
And $$ H^2 = \frac{k_1^2+ k_2^2 +2k_1k_2 }{2} \geq \frac{|K| +
   K}{2} \geq K $$