Curvature of Fernet curve on a sphere The question is, how to prove that the curvature of any Frenet curve on a sphere with radius $R$ is bigger or equal to $1/R$.
I have managed to prove so far that the Gauss curvature of the sphere $x^2+y^2+z^2=R^2$ is $1/R^2$, but I don't know if this helps at all
 A: Suppose $\alpha(s)$ is a unit speed curve lying in the sphere of radius $R$ centered at the origin.  Then
$\alpha(s) \cdot \alpha (s) = R^2, \tag 1$
whence
$\dot \alpha(s) \cdot \alpha(s) = 0; \tag 2$
since
$\dot \alpha(s) = T(s), \tag 3$
the unit tangent vector to $\alpha(s)$, (2) becomes
$T(s) \cdot \alpha (s) = 0; \tag 4$
differentiating this equation yields
$\dot T(s) \cdot \alpha(s) + T(s) \cdot \dot \alpha(s)  = 0; \tag 5$
we now recall (3), viz.
$\dot \alpha(s) = T(s) \tag 6$
and the Frenet-Serret equation
$\dot T(s) = \kappa(s) N(s); \tag 7$
then (5) yields
$\kappa(s) N(s) \cdot \alpha(s) + T(s)  \cdot T(s) = 0; \tag 8$
also,
$T(s) \cdot T(s) = 1, \tag 9$
$T(s)$ being a unit vector.  (8) may now be written
$\kappa(s) N(s) \cdot \alpha(s) = -1; \tag{10}$
note this forces
$\kappa(s) \ne 0; \tag{11}$
taking absolute values in (10) we find
$\kappa(s) \vert N(s) \cdot \alpha(s) \vert = 1; \tag{12}$
by Cauchy-Schwarz,
$ \vert N(s) \cdot \alpha(s) \vert  \le \vert \alpha(s) \vert \vert N(s) \vert = R, \tag{13}$
since
$\vert \alpha(s) \vert = R \tag{14}$
and
$\vert N(s) \vert  = 1; \tag{15}$
assembling (12) and (13) together we have
$\kappa(s) R \ge 1, \tag{16}$
or
$\kappa(s) \ge \dfrac{1}{R}, \tag{17}$
$OE\Delta$.
