# Surface area of a regular, orientable, compact surface with positive Gaussian curvature

Problem: Let $S$ be a regular, orientable, compact surface with positive Gaussian curvature: $K ≥ M > 0$. Prove that the surface area of $S$ is less than $\dfrac{4\pi}{M}$

Ideas: From the Gauss-Bonnet Theorem, we have that $$\int_S K=2\pi\chi(S)$$ As a corollary, we have that if $K>0$, then $S$ is homeomorphic to a sphere, which has $$\int_SK=2\pi\chi(S)=4\pi$$ However, I am not quite sure how to sure that the surface area would be less than $\dfrac{4\pi}{M}$. What other ideas do I need?

The surface area of $S$ is given by $\int_S 1$, and
$$4\pi = 2\pi\chi(S) = \int_S K \geq \int_S M = M\int_S 1.$$