Curvature of a curve on a sphere I'd like some hint for this problem:
Show that every unit-speed curve $ \alpha : I \subset \mathbb{R} \longrightarrow \mathbb{R}^3$, whose image is in the sphere of radius $R$ , has curvature $\displaystyle k_{\alpha}(s)\geq \frac{1}{R}$ .
 A: $\alpha(t) \equiv (x(t), y(t), z(t))$ 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
$$k_\alpha R = \Vert (\ddot x, \ddot y, \ddot z) \Vert R \ge 1$$
A: 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.
A: 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.
