Characterize the sphere using mean curvature.

Question

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!

Answer 1

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 $$ H^2 = \frac{k_1^2+ k_2^2 +2k_1k_2 }{2} \geq \frac{|K| + K}{2} \geq K $$