Any compact embedded $2$-dimensional hypersurface in $\mathbb R^3$ has a point of positive Gaussian curvature Problem statement: Let $M \subseteq \mathbb{R}^3$ be a compact, embedded, 2-dimensional Riemannian submanifold. Show that $M$ cannot have $K \leq 0$ everywhere, where $K$ stands for the Gauss curvature of $M$.
I have attempted an approaches (described below), but have not been able to finish it off.
I came across a hint suggesting that I consider the square-of-distance function $f: \mathbb{R}^3 \rightarrow [0,\infty)$ by $f(x)=|x|^2$. As $M$ is compact, there is some $q_0 \in M$ such that
\begin{equation}
\forall q \in M: \quad f(q) \leq f(q_0).
\end{equation}
We also know that
\begin{equation}
\forall v \in T_{q_0} M: \quad df_{q_0}(v) = \frac{d}{dt}\bigg|_{t=0} (f \circ \gamma_v)(t) = 0\quad \text{and} \quad (f \circ \gamma_v)''(0) \leq 0.
\end{equation}
The problem now is that I do not know how to relate the above statements to a statement about the curvature $K(q_0)$.
I would appreciate any suggestions on either of the above approaches.
 A: This is not a suggestion on your approach, but rather a different approach.
As $M$ is compact, there exists a smallest sphere $S$ containing $M$, and it must touch $M$ at some point $p$. Hence the Gaussian curvature of $M$ at $p$ is at least as big as the Gaussian curvature of $p$ at $S$, which is positive, hence it is itself positive.
A: That's very nice. Approach 2 works.  You are calling the farthest point $q_0.$ There is a method called Lagrange multipliers. The tangent plane to your surface is orthogonal to the vector from the origin to $q_0.$ Meanwhile, your surface is contained in the closed ball around the origin that passes through $q_0.$ As a result, the principal curvatures are nonzero and have the same $\pm$ sign as the sphere containing $M.$ (If not, there would be points arbitrarily close to $q_0,$ along a principal direction, outside the closed ball, because they would be in the tangent plane at $q_0,$ or between the tangent plane and the tangent sphere, or on the wrong side of the tangent plane). So, the two principal curvatures are nonzero and have the same $\pm$ sign, so their product $K$ is strictly positive.
See http://en.wikipedia.org/wiki/Principal_curvature 
