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## Tangent surfaces curvature inequality [closed]

I found this lemma in a few surface geometry proofs:

If we have two surfaces, $S$ and $S'$, which are tangent in the point $p$ then if: (i) $S'$ has positive curvature in $p$; (ii) $S$ is, locally around $p$, situated on the same side of $S'$, then the curvature of $S$ in $p$ is greater or equal to the curvature of $S'$ in $p$.

I am interested in a book/reference where I can find a proof for this lemma. Thank you.

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What does "on the same side" mean? The way I read this, "on the same side" seems to me to be a symmetric relation. Yet you somehow derive an inequality from it. – José Figueroa-O'Farrill Feb 20 2011 at 12:06
Ok. I will edit it. I didn't realize that it was wrong in this form. I wanted to express that $S$ and $S'$ do not intersect, and $S$ is somewhat "inside" $S'$, since $S'$ has positive curvature at $p$. Please enlighten me how to express that in a more rigorous way... – Beni Bogosel Feb 20 2011 at 12:36
Why don't you simply quote the precise lemma from one of the sources you have read this from? – José Figueroa-O'Farrill Feb 20 2011 at 13:13
My guess is that you mean the lemma which is used, for example, in the proof of the fact that a bounded surface in $\mathbb{R}^3$ has a point of positive curvature. One proves this by bringing a sphere out from infinity until it just touches the surface and at that point you are in the situation you are describing. Then the lemma says that at that point the curvature of the surface is bounded below by that of the osculating sphere. This is a simple lemma using the local form of a surface around a point of positive curvature and you should be able to find this in any number of textbooks. – José Figueroa-O'Farrill Feb 20 2011 at 13:26
Take the tangent plane P at p to S and S'. You can locally express the surfaces as graphs over P using the implicit function theorem. The Gaussian curvatures at p are now just the determinant of the Hessians of the functions representing S and S' (the gradient vanishes since P is tangent). Now it is just an exercise in Taylor's theorem to show that if the graphs $S \geq S'$ in a neighborhood, then as a quadratic form the hessian of S $\geq$ that of S', and hence the determinant also. Like José said: you should be able to find this in most textbooks on differential geometry of surfaces. – Willie Wong Feb 20 2011 at 14:00