An inequality for absolute total curvature in Riemannian surfaces

Question

Let be $M\subseteq \mathbb{R}^3$ a compact (Riemannian) surface and let be $K$ the gaussian curvature of $M$.

I want to prove that $$ \int_{M} |K| \geq 4\pi(1+g(M))$$ where $g(M)$ is the genus of $M$. I tried using the following inequalities:

$$ \int _M |K|=\int_{K>0} K - \int_{K<0} K \geq 4\pi -\int_{K<0} K\geq 4\pi -\int_M K$$

but I had no luck even using Gauss Bonnet. Any hint?

Answer 1

You are almost there, you just throw away too much:

$$\begin{split} \int_M |K| &= \int_{K>0} K - \int_{K<0} K = \int_{K>0} K - \left(\int_{K<0} K + \int_{K>0} K - \int_{K>0} K\right)\\ &= \int_{K>0} K - \left(\int_M K - \int_{K>0}K \right)\\ &= 2\int_{K>0} K - \int_M K\ . \end{split}$$