Travelling on a sphere A particle travels on a unit sphere from a point $(z, \theta) = (z_0, 0)$ to $(z_0, \theta_0)$ in cylindrical coordinates, so z0 = const. Find the distance travelled and compare with the "great circle" (G.C.) distance between those points.
My solution:
a. $Distance = r  \theta = \sqrt{(1 - z_0^2)}  \theta_0$, where r is the radius of a circle, cut from sphere by $z = z_0$.
b. The great circle arc distance can be found by a dot product of vectors to the beginning and the end of the journey:
$r_0 = (\sqrt{(1 - z_0^2)}, 0, z_0)$
$r1 = (\sqrt {(1 - z_0^2)} \cos \theta_0, \sqrt {(1 - z_0^2)} \sin \theta_0, z0)$
$G.C. Distance = \arccos (r_0 r_1) = \arccos((1-z_0^2) \cos \theta_0 + z_0^2) $
My question: how to show (from this calculation that $Distance >= G.C. Distance$. I can do it from geometric considerations, but would like to derive it analytically from the formulas above (or any alternative formulas that can be derived).
 A: You want to show that
$$
\arccos[(1-z_0^2)\cos\theta_0+z_0^2]\le\theta_0\sqrt{1-z_0^2},
$$
for every $\theta_0\in[0,\pi]$. As the cosine is decreasing on $[0,\pi]$, that is the same as proving that
$$
(1-z_0^2)\cos\theta_0+z_0^2\ge\cos\left(\theta_0\sqrt{1-z_0^2}\right).
$$
If we define
$$
f(\theta)=(1-z_0^2)\cos\theta+z_0^2-\cos\left(\theta\sqrt{1-z_0^2}\right)
$$
we must then prove that $f(\theta)\ge0$ for $\theta\in[0,\pi]$.
Notice that $f(0)=0$ and $f'(0)=0$, but on the other hand
$$
f''(\theta)=(1-z_0^2)\left[\cos\left(\theta\sqrt{1-z_0^2}\right)-\cos\theta\right]\ge0
$$
for $\theta\in[0,\pi]$. It follows that $f'(\theta)$ is increasing, that is $f'(\theta)\ge0$ for $\theta\in[0,\pi]$, which in turn implies that $f(\theta)$ is also increasing, that is $f(\theta)\ge0$ for $\theta\in[0,\pi]$, as we were to prove.
A: Just for the sake of completeness, here is my proof that Distance ($d$) is longer than G.C.Distance ($D$). Both trajectories are the arcs of the circles (of radius $r = \sqrt{(1-z_0^2)}$ and $ R = 1$ respectively) around the chord $c$. The corresponding central angles are $\theta$ and $\phi$ respectively. 
It is easy to see that $c/2 = r \sin{(\theta/2)} = R \sin{(\phi/2)}$, so 
$\sin{(\phi/2)}/ \sin{(\theta/2)} = r/R \le 1$. 
Since $\sin(x/2)$ monotonically increases between $0$ and $\pi$, $\phi \le \theta$.
Now let's compare the arc lengths, $D = R\phi$ and $d = r\theta$:
$D/d = R/r * \phi/\theta = \sin (\theta/2) / \sin (\phi/2) * (\phi/2) / (\theta$/2)
Since $f(x) = \sin(x)/x$ monotonically decreasing from $0$ to $\pi$, $sin(\theta/2)/(\theta/2) \le sin(\phi/2)/(\phi/2)$ and
$D/d \le 1$.
