A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency Suppose $a$ is an arc length-parametrized space curve with the property that 
$\|a(s)\| \leq \|a(s_0)\| = R$ for all $s$ sufficiently close to $s_0$. Prove that 
$k(s_0) \geq 1/R$. 
So, I was going to consider the function $f(s) = \|a(s)\|^2$ but then the hint asks what we know about $f''(s_0)$? I am confused on how to link the two together and show from there that the curvature of $k(s_0) \geq 1/R$. 
I know that we should use the second derivative test to see the local minima or maxima thus giving us if the curve points upward or downward, but I am a little lost at where to begin. I was told that $f(s_0)$ is constant, why? How come $f(s)$ is not a constant function here? 

Here's what I think is going on, I'll let you fill in the details.
The relevant definitions are all included here: http://en.wikipedia.org/wiki/Curvature#Curvature_of_space_curves
Because $a$ is parametrized by the arc-length, the curvature at a point $s$ is simply $\kappa(s) = ||a''(s)||$. Now imagine a curve $\gamma$ (also parametrized by the arc-length) whose image is the intersection of a sphere of radius $R$, and a plane passing through the origin in $\mathbb{R}^3$, and such that $\gamma(s_{0}) = a(s_{0})$ (and if you want $\gamma'(s_{0})= a'(s_{0})$); in this case the curvature of $\gamma$ at $s_{0}$ (and at any other point for that matter) is simply $\kappa(s_{0}) = 1/R$. 
Now we know that $||a(s)|| \leq ||a(s_{0})|| = R$ for $s$ close to $s_{0}$. If you think about this geometrically, this means that near $s_{0}$, the curve $a$ does not leave the sphere of radius $R$; in fact it may even curve inward. Therefore, it is not too hard to see that $||a''(s_{0})|| \geq ||\gamma''(s_{0})||$, or in other words, $\kappa(s_{0}) \geq 1/R$.
