Why must this function have a critical point inside the sphere? Suppose we have $f: \mathbf{R}^{3} \to \mathbf{R}$ with the following property: $\langle \nabla f(x), x \rangle > 0$ for every $x \in S^{2}$, that is, it's gradient points outwards the unit sphere. It's asserted that there must a point $p$ inside the sphere with the property $\nabla f (p) = 0$.
Here's what I've done so far: suppose there's no such point in $B(0;1)$. Since $f$ is real valued and defined in the compact $\bar{B}(0;1)$, it must attain maximum and minimum. Since $f$ has no critical points in the interior, then these points must lie in $S^{2}$, say $x_{0}$ is the maximum and $y_{0}$ is the minimum. Now the problem seems to be that the gradient cannot point outwards in the minimum point, and from that we could derive a contradiction. But I don't know how to write that down -- using the directional derivative along the line joining the extrema points perhaps?
 A: You have it. At each point $x$ of the sphere, the directional derivative of $f$ at $x$ in the normal direction is positive, which says that $f$ increases as you approach $x$ from inside the ball. This means that the global minimum of $f$ on the closed ball cannot occur at a boundary point. 
A: Here is a proof that appeals to the Poincare-Hopf theorem:

If $M$ is a manifold with boundary and $X$ is a vector field with isolated zeroes on $M$ and pointing outward on $\partial M$, then 
  $$\sum_{\mbox{zeroes }m_i} \operatorname{index}_{m_i}(X) = \chi(M).$$

Since $\nabla f$ is a smooth vector field on the closed unit ball and the Euler characteristic of the closed unit ball is $1$, there must be at least one point in the interior of $M$ for which $\nabla f = 0$.
A: There is a topological argument. Consider the map
$$\sigma : S^2 \to S^2 : x \mapsto \frac{\nabla f}{\|\nabla f\|}.$$
By the condition on $f$ at the boundary, $\sigma$ is homotopic to the identity map of $S^2$ (because there is a unique geodesic going from $x$ to $\sigma(x)$). Now suppose that $\|\nabla f\| \neq 0$ for each $x \in \overline{B}(0,1)$. Consider the map
$$\overline{B}(0,1) \to S^2 : x \mapsto \frac{\nabla f}{\|\nabla f\|}.$$
Since $\overline{B}(0,1)$ is contractible, this map is homotopic to a constant map. But that is absurd, since its restriction to the boundary $S^2$ is $\sigma$, which is not homotopic to a constant map.
Cool, huh?!
