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? 

Some intermediate hints:


*

*We are given that $\Vert a(s) \Vert \leq \Vert a(s_0)\Vert$ for all $s$ near $s_0$.  Squaring both sides tells us that $f(s) \leq f(s_0)$ for all $s$ near $s_0$. In other words, $s = s_0$ is a local max for the function $f$.  What does that imply about $f''(s_0)$?

*You should compute $f''(s_0)$ as the hint says.  This means that you'll have to compute $f'(s)$, then its derivative $f''(s)$, then finally plug in $f''(s_0)$.  To compute these derivatives, you should use that $f(s) = \Vert a(s) \Vert^2 = \langle a(s), a(s) \rangle$, and use the product rule for inner products:
$$\frac{d}{ds}\langle F(s), G(s) \rangle = \langle F'(s), G(s) \rangle + \langle F(s), G'(s) \rangle.$$

*Finally, once you've computed $f''(s)$ and also used Hint 1, you should get an inequality.  At this point, you might want to apply the Cauchy-Schwarz Inequality
$$\left|\langle v,w \rangle\right| \leq \Vert v \Vert \Vert w \Vert.$$
You'll also want to remember that since $a(s)$ is unit speed, we have $\Vert a'(s) \Vert = 1$ for all $s$.
