Prove that the curvature of $\alpha$ is greater than $1/R^2$ for $\alpha$ lying on a sphere I want to prove that any $\alpha(s)$ with arc parameter $s$ lying on a sphere with radius $R$ has curvature $k\geq1/R^2$.
I tried so many times to prove it and at this point it's pissing me off really hard. The conditions of arc length parameter $$ |\alpha'(s)|=1 $$
and lying in the sphere $$|\alpha(s)|=R$$ should be enough to prove this by deriving those equations etc.
The only thing I have found with this is that $$ \{\alpha,\alpha',\alpha\wedge \alpha' \} $$ is a basis of the curve at each point. This works for nothing to obtain the curvature $k$, as curvature is defined as
$$k(s) = |\alpha''(s)| $$
or as the function such that
$$ \alpha''(s) = k(s)N(s)$$ where $N$ is in a vector in the direction of $\alpha''$.
I also tried methods such as starting first with plane curves of maximum radius, which gives me a curvature of $1/R^2$. Then saying that plane curves of less radius have curvature $1/\tilde{R}^2\geq 1/R^2$ as $\tilde{R}$ is smaller. But I don't have any clue for the other type of curves in the sphere.
Note: I have seen many similar posts but none address this problem but other properties about the curvature of a curve in a sphere.
 A: I think I found a solution without using unknown concepts such as geodesics and geodesic curvature and will post the answer to my own question.
As $\alpha$ is lying on the sphere of radius R, $|\alpha(s)|=R$ and deriving the expression we obtain $\langle \alpha, \alpha' \rangle=0$.
Using the Frenet frame, we know that $\alpha = aT + bN + cB$. As $\langle \alpha, \alpha' \rangle=0 \implies a = 0$ so $\alpha = bN + cB$.
Deriving this and using the Frenet formulas we see that
$$ \alpha' = -b(kT-\tau B) + c\tau N $$
As $\alpha$ is parametrised by arc length, $-bk=1$ and using that $|b|\leq R$ and taking absolute value,
$$k=\frac{1}{|b|}\geq\frac{1}{R} $$ qed
A: Let us denote the dot product of two vectors $\vec v$ and $\vec u$ by $\langle \vec v, \vec u \rangle$.
Parameterize the curve $\alpha = \alpha(s)$ with arc length $s$. Then, $|\alpha'(s)| = 1$. Because $\alpha$ is a part of the sphere, $|\alpha(s)| = R$. Differentiating $R^2 = \langle \alpha(s), \alpha(s) \rangle$, we obtain $2 \cdot \langle \alpha(s), \alpha'(s) \rangle = 0$, $\implies \alpha(s) \perp \alpha'(s)$. Differentiating $\langle \alpha(s), \alpha'(s) \rangle$ further, we get $\langle \alpha(s), \alpha''(s) \rangle$ $+ \langle \alpha'(s), \alpha'(s) \rangle$ $=0$. But, $|\alpha'(s)| = 1$ $\implies$ $\langle \alpha'(s), \alpha'(s) \rangle = 1$.
So, $\langle \alpha(s), \alpha''(s) \rangle$ $=-1$.
Now, notice that $|\langle \vec v, \vec u \rangle | \leq |\vec v| \cdot |\vec u|$. Therefore, $1 = |\langle \alpha(s), \alpha''(s) \rangle | \leq |\alpha(s)| \cdot |\alpha''(s)|$ $= R \cdot |\alpha''(s)|$.
So, $$|\alpha''(s)| \geq \dfrac{1}{R}$$
$\blacksquare$.
