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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?

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    $\begingroup$ Welcome to Math.SE! What is $K$? Please include your definition of $K$ in the question. $\endgroup$
    – Hrodelbert
    Commented Jun 8, 2015 at 9:10
  • $\begingroup$ I explained, thank you for pointing out! $\endgroup$
    – Manticora
    Commented Jun 8, 2015 at 9:13

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

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

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  • $\begingroup$ @John, I don't see what is happening in your last equality. Can you explain how you end up with that? $\endgroup$
    – Hrodelbert
    Commented Jun 8, 2015 at 9:29
  • $\begingroup$ I added one more line, is it okay now? @Hrodelbert $\endgroup$
    – user99914
    Commented Jun 8, 2015 at 9:31
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    $\begingroup$ @John, That was embarrassingly easy: for some reason I had not spotted the crucial minus sign. Thanks anyway! $\endgroup$
    – Hrodelbert
    Commented Jun 8, 2015 at 9:33

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