Curve on a sphere

Regard $\alpha\colon I \to \mathbb{S}^2_R$ as a map from the interval $I$ to the sphere of radius $R$. Let $\langle \cdot, \cdot \rangle$ denote the dot product. Here are two hints:

(1) Consider the quantity $\frac{d}{ds}\langle \alpha(s), \alpha'(s)\rangle$. What do you know about the quantity $\langle \alpha(s), \alpha'(s)\rangle$ based on the geometry of the sphere?

(2) Since you're trying to prove an inequality, perhaps you know of some inequality involving the inner product $\langle v, w\rangle$ of two vectors.

$\alpha(t) \equiv (x(t), y(t), z(t))$ verifies the identity $$x^2 + y^2 + z^2 = R^2$$ deriving twice the previous identity we deduce $$\left\vert x\ddot x + y\ddot y + z\ddot z \right\vert = 1$$ and applying the Cauchy–Schwarz inequality to the left side we obtain $$k_\alpha R = \Vert (\ddot x, \ddot y, \ddot z) \Vert R \ge 1$$

Let's see now . . . this question was posted on Wednesday 29 June 2011, more than eight years ago. I suppose then, that our OP Jr. has had more than enough time to digest the hints he requested. Therefore, for those who wish to see the details, I present:

We assume $\alpha(s)$ is a unit-speed curve lying in the sphere of radius $R$ centered at the point $c \in \Bbb R^3$; then $\alpha(s)$ satisfies

$(\alpha(s) - c) \cdot (\alpha(s) - c) = R^2; \tag 1$

we differentiate this equation with respect to $s$, and obtain

$\dot \alpha(s) \cdot (\alpha(s) - c) = 0; \tag 2$

since $\alpha(s)$ is a unit-speed curve, it has a unit tangent vector

$T(s) = \dot \alpha(s), \; T(s) \cdot T(s) = 1, \tag{2.5}$

and thus we write (2) in the form

$T(s) \cdot (\alpha(s) - c) = 0, \tag 3$

which we may differentiate yet again with respect to the arc-length $s$:

$\dot T(s) \cdot (\alpha(s) - c) + T(s) \cdot \dot \alpha(s) = 0; \tag 4$

according to (2.5) this becomes

$\dot T(s) \cdot (\alpha(s) - c) + 1 = 0, \tag{4.1}$

which clearly implies

$\dot T(s) \ne 0; \tag{4.2}$

thus we may next introduce the Frenet-Serret relation

$\dot T(s) = \kappa(s) N(s), \; \kappa(s) > 0, \vert N(s) \vert = 1, \tag{4.5}$

and substitute it into (4.1):

$\kappa(s)N(s) \cdot (\alpha(s) - c) + 1 = 0, \tag 5$

now

$\alpha(s) - c = R \mathbf n(s), \tag 6$

where $\mathbf n(s)$ is the outward-pointing unit normal vector to the sphere at the point $\alpha(s)$; thus

$\kappa(s) R N(s) \cdot \mathbf n(s) + 1 = 0; \tag 7$

we note this formula forces

$N(s) \cdot \mathbf n(s) \ne 0, \tag 8$

and therefore we may write

$\kappa(s) = - \dfrac{1}{RN(s) \cdot \mathbf n(s)}, \tag 9$

whence, since $\kappa(s) > 0$,

$\kappa(s) = \vert \kappa(s) \vert = \dfrac{1}{\vert R N(s) \cdot \mathbf n(s) \vert} = \dfrac{1}{R \vert N(s) \cdot \mathbf n(s) \vert}; \tag{10}$

now $N(s)$ and $\mathbf n(s)$ being unit vectors it follows in light of (8) that

$0 < \vert N(s) \cdot \mathbf n(s) \vert \le \vert N(s) \vert \vert \mathbf n(s) \vert = 1; \tag{11}$

therefore,

$\dfrac{1}{\vert N(s) \cdot \mathbf n(s) \vert} \ge 1, \tag{12}$

and finally,

$\kappa(s) = \dfrac{1}{R \vert N(s) \cdot \mathbf n(s) \vert} = \dfrac{1}{\vert N(s) \cdot \mathbf n(s) \vert } \dfrac{1}{R} \ge \dfrac{1}{R}. \tag{13}$

In closing note that from (7) we have

$N(s) \cdot \mathbf n(s) = -\dfrac{1}{\kappa(s) R} < 0, \tag{14}$

which shows that $N(s)$ is always pointing towards the interior of the sphere.