Let $C \subset \mathbb{C}$ be a (locally rectifiable) Jordan curve separating the pairs of points $a_1,b_1$ and $a_2,b_2$ (i.e. $a_1,b_1$ lies in $\text{ins}(C)$ and $a_2,b_2$ lies in $\text{out}(C)$, each pair of points lies in a different component of $C$'s complement). Moreover assume that $k(a_i,b_i) \geq \delta > 0, i=1,2$. Here $k(x,y)$ denotes the spherical length between $x$ and $y$, that is, $k(x,y) = \inf \int_{\gamma} (1+|z|^2)^{-1} |dz|$ where the infimum is taken over all locally rectifiable Jordan arcs joining $x$ and $y$. (This is the usual notion of spherical length, i.e. geodesics are great circles).
Then apparently by "elementary geometric considerations", one can immediately deduce that the spherical length of $C$, $\ell_{k}(C)$ satisfies $\ell_{k}(C) \geq 2 \delta$.
One can replace 'locally rectifiable' in the above by any class of sufficiently regular Jordan arcs, e.g. $C^{\infty}$ smooth Jordan arcs.
I am likely missing something very obvious here, in any case I would be grateful for any help.
$\textbf{Speculation}$: I am guessing one way to think about this is that we can extend the (smaller) geodesic segment $[a_1,b_1]$ and look at the arcs subtended at the intersection points of the segment with $C$, these should both have length $\geq \delta$... I'm just not sure if this really works for arbitrary Jordan curves (and if it does work, I don't understand why we need two pairs of points).