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Let $N$ be a hypersurface in $\mathbb R^n$, assume it is compact. Then the maximum point of $d(O, x)$ when restrict to $N$ has positive sectional curvature lower bound by the one of the correspond sphere tangent to it, right?

The best lower bound would be the smallest sphere contains $N$, right?

I am asking this since it seems someone estimates the Ricci curvature of such point, a natural question is why not estimate the sectional curvature if what I said is correct?

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    $\begingroup$ No, the spheres estimate principal curvatures. In $\mathbb R^3$ the sectional curvature equals Gauss curvature, which is a single number at a given point, but there are two principal curvatures which correspond to the spheres. $\endgroup$
    – Sergei Ivanov
    Commented Mar 23, 2013 at 17:16
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    $\begingroup$ @Sergei, the sectional curvatures at the furthest point do have positive sign right? Since it is locally convex and 'supported' by round sphere with certain radius. @Unknown, I think you need to rephrase your question to make it clear. $\endgroup$
    – Ralph
    Commented Mar 23, 2013 at 19:35
  • $\begingroup$ @Ralph, yes, probably you read the question better than I did. I reacted to the phrase 'best lower bound', which I read as 'the minimum sectional curvature at a point'. $\endgroup$
    – Sergei Ivanov
    Commented Mar 23, 2013 at 20:10

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If the extrinsic diameter of $N$ in $\mathbb{R}^n$ is $D = 2/\rho$ where $\rho>0$, then for any two points $p$ and $q$ in $N$ that are $D$ units apart, all the sectional curvatures at $p$ and $q$ are are at least $\rho^2$, so the Ricci curvature (in any tangent direction) at those two points will be at least $(n{-}1)\rho^2$.

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