Let $\gamma: (\alpha, \beta) \rightarrow \mathbb R^3$ be a space curve with speed $1$ s.t $|| \gamma(s) || = R > 0$.
I want to show that the curvature $\kappa(s) \ge \frac 1 R$ for all $s \in (\alpha, \beta)$.
I've tried various several things among them $\frac {d} {ds} ||\gamma(s)||$, however I don't get foothold to conclude anything.
Let $\gamma$ be given in the form $s\mapsto x(s)$. From $\bigl|x(s)\bigr|\equiv R$ we obtain $x\cdot\dot x=0$, and differentiating again produces $\dot x\cdot\dot x+x\cdot\ddot x=0$, or $x\cdot \ddot x=-1$. As $|x|=R$ we obtain $\kappa:=|\ddot x|\geq{1\over R}$, by Schwarz' inequality.
If this was too fast we can argue as follows:
You may assume that the curve $\gamma$ is lying on the unit sphere $S^2$. Let $P={\bf x}(0)$ be an arbitrary point on $\gamma$. You can then move $\gamma$ rigidly around the sphere so that $P$ becomes the point $(1,0,0)$ on the equator, and $\dot{\bf x}(0)$ points in the direction of the north pole. In this way $\gamma$ has a parametric representation of the form $$\gamma:\quad s\mapsto{\bf x}(s):=\bigl(\cos\phi(s)\cos\theta(s),\sin\phi(s)\cos\theta(s),\sin\theta(s)\bigr)$$ with $\phi(0)=\dot\phi(0)=\theta(0)=0$. One computes $$|\dot{\bf x}(s)|^2=\cos^2\theta(s)\dot \phi^2(s)+\dot\theta^2(s)\ ,$$ and this implies $\dot\theta(0)=1$. A lengthy computation (which I left over to Mathematica) gave $$\ddot{\bf x}(0)=\bigl(-\dot\phi^2(0)-\dot\theta^2(0),\ddot\phi(0),\ddot\theta(0)\bigr)= \bigl(-1,\ddot\phi(0),\ddot\theta(0)\bigr)\ ,$$ so that we obtain $$\kappa(0)=\bigl|\ddot{\bf x}(0)\bigr|=\sqrt{1+\ddot\phi^2(0)+\ddot\theta^2(0)}\geq1\ .$$