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If $M$ is a smooth closed $n$-dimensional Riemannian manifold which is Riemannian embedded in $\mathbb R^{n+1}$, then there exists a point $p \in M$ such that the sectional curvatures at $p$ are all positive.

Can any one give me a hint for this problem? I was considering the maximum $p$ of function $|x|^2$ on $M$, then near $p$, $M$ is "wrapped" by some $S^n$ and has the same tangent space as $S^n$. But I am stuck there.

I have made some progress:

We consider the functions $L_q(x)=|x-q|^2$. Then we have a maximum $p$ of $L_q$ and we fix the unit vector $v=\frac{p-q}{|p-q|}$ throughout, so $v$ is the normal vector at $p$. Now if we set $q(t)=p+tv$, then when $t\leq-|p-q|$, $p$ is always the maximum of function $L_{q(t)}$ (this is true if we draw a ball at $q(t)$ with radius $|p-q(t)|$, then all $M$ is contained in this ball).

Therefore when $t$ sufficiently tends to $-\infty$, the Hessian $L_{q(t)}$ is always semi-positive definite. Now if we fix a coordinate neighborhood aroud $p$, then Hessian matrix $H$ of $L_{q(t)}$ at $p$ is given by $H=2(F-tS)$ where $F,S$ are the first and second fundamental forms of $M$ at $p$. So we conclude that $S$ has to be semi-positive definite.

But how can we move further to say $S$ is positive definite?

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    There is a trivial counter example, namely, $S^1\subseteq \mathbb{R}^2$. Perhaps you're assuming $n>1$? Also, for the $n=2$ case, see http://math.stackexchange.com/questions/89061/are-there-any-compact-embedded-2-dimensional-surfaces-in-mathbb-r3-that-are/89073#89073.2012-10-20
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    @Hezudao: concerning your edit (partial progress): my answer solves the problem along these lines, except that it's simpler (using linear functions rather than distance squared) - and complete :)2012-10-24

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