An exercise in the Riemannian geometry book 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?
 A: There exists a linear function $L:\mathbb R^{n+1}\to\mathbb R$ such that $L|_M$ has non-degenerate critical points (by Sard's theorem). If you choose a maximum of $L$ on $M$ then  the second quadratic form at that point is positively definite, hence all the sectional curvatures at that point are positive.
edit (why positive 2nd fund. form implies positive curvature):
if $K$ is the 2nd fundamental form then the curvature tensor is given by
$$(R(u,v)u,v)=K(u,u)K(v,v)-K(u,v)^2$$
and if $u,v$ are linearly independent and $K$ is positively definite then the RHS is positive. Sectional curvatures are the LHS if $(u,u)=(v,v)=1$ and $(u,v)=0$.
A: One way to obtain the Levi-Civita connection for embedded submanifolds in $\mathbb{R}^{n+1}$ is to take the covariant derivative of a vector field with respect to the standard flat connection in $\mathbb{R}^{n+1}$, and then project the resulting vector into the tangent space. Thus the covariant derivative of a curve through the point of tangency on the manifold matches that of the sphere. Consequently, so does the Riemann curvature. (I'm a little fuzzy on this step: do I need the vector fields to be equal in a neighborhood, instead of just at a point?). So at the point of tangency, the sectional curvatures of the manifold match those of the sphere.
