Is it true that for every differentiable curve $\gamma:(a,b)\to S^2$ parametrized by arc length, we have $\|\gamma''(t)\| \geq 1$ for every $t\in (a,b)$?.
I want to know if this is correct:
Let $II_p$ denote the second fundamental form at the point $p$ and $N_p$ the Gauss map at the point $p$. Then we have $dN_p = Id$ for every $p$ in $S^2$ since the surface we consider is $S^2$. Therefore, for $t \in (a, b)$, we have $-1 = -\|\gamma'(t)\|^2 = -\langle \gamma'(t), \gamma'(t)\rangle = -\langle dN_{\gamma(t)}[\gamma'(t)], \gamma'(t)\rangle = II_{\gamma(t)}[\gamma'(t)] = \langle N(\gamma(t)), \gamma''(t)\rangle$.
Thus, $1 = |\langle N(\gamma(t)), \gamma''(t)\rangle| \leq \|N(\gamma(t))\| \cdot \|\gamma''(t)\| = \|\gamma''(t)\|$, where I used the Cauchy-Schwarz inequality and $\|N(\gamma(t))\| = 1$.
There is a very direct proof using only basic calculus:
From $\|\gamma\|^2=1$ we get by differentiating $0=\gamma'\cdot \gamma$ and $0=\gamma''\cdot\gamma+\|\gamma'\|^2\,.$ Since $\gamma$ is parametrized by arc length we know that $\|\gamma'\|=1\,.$ So we have $$ |\gamma''\cdot\gamma|=1\,. $$ By the Cauchy-Schwarz inequality $$ |\gamma''\cdot\gamma|^2\le \|\gamma''\|\underbrace{\|\gamma\|}_{=1} $$ and your claim follows.