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**Total "absolute" curvature of a compact surface**

Hi! Someone could help me resolving the following problem? Let [tex]\Sigma \subset \mathbb{R}^3[/tex] be a compact surface: show that

[tex]

\int_{\Sigma}{|K|\mathrm{d}\nu} \ge 4\pi

[/tex]

where [tex]K[/tex] is the gaussian curvature of [tex]\Sigma[/tex]. The real point is that I want to prove this using weaker results as possible (in particular, the Gauss-Bonnet theorem). I tried using that

[tex]

\int_{\Sigma}{|K|\mathrm{d}\nu} = \text{Area}\left(N(\Sigma^+)\right) + \text{Area}(N(\Sigma^-))

[/tex]

where

[tex]\Sigma^{\pm}:=\left\{p \in \Sigma \biggr| \text{sign}\left(K(p)\right)=\pm 1\right\}[/tex]

[tex]N[/tex] is the Gauss application

and the formula is to be understood as "counted with multiplicity", i.e. we don't consider overlaps

(as follows from, e.g., the proof of Proposition 2 on page 167 of do Carmo's Differential Geometry). I'm pretty sure that this way will bring to some result, but i don't have any idea how to proceed from here! Thanks in advance for your help!