Equivalence of chordal and spherical metric on Riemann Sphere The chordal metric between two points $z,w \in \hat{\mathbb{C}} (=\mathbb{C}\cup{\{\infty\}})$ is defined as ,
\begin{align*}
    d(z,w)=\displaystyle\frac{2|z-w|}{\sqrt{(1+|z|^2)(1+|w|^2)}}
 \end{align*}
\begin{align*}
     d(z,\infty)=\displaystyle\frac{2}{\sqrt{(1+|z|^2)}}
 \end{align*}
The spherical metric is defined as goes ,
$\gamma \colon [0,1] \to \mathbb{C}\cup{\{\infty\}}$; The spherical length of the curve $\gamma$ in  $ \hat{\mathbb{C}} $ is defined as,
\begin{align*}
    \Lambda(\gamma)= \int_\gamma ds = \int_\gamma \displaystyle\frac{2|dz|}{1+|z|^2}
\end{align*}
The spherical distance between two points is  the infimum of the spherical lengths of paths joining the two points:
\begin{align*}
     \rho (z,w) := inf_\gamma \Lambda(\gamma)
 \end{align*}
$\gamma$ varies over all paths joining $z$ and $w$.
In "Normal Families" by schiff, it's written that ,
\begin{align*}
     d(z,w) \leq \rho(z,w) \leq \dfrac{\pi}{2} d(z,w)
 \end{align*}
I cannot figure out why is this true .
 A: Let me first introduce my notations $d_c(z,w)=$ Chordal distance between $z$ and $w$. $d_s(z,w)=$ Spherical distance between $z$ and $w$ and $S=$ Unit sphere in $\mathbb{R^3}$. Now suppose that $z$ and $w$ maps to $Z$ and $W$ in the sphere under the stereographic map.
Now I can always intersect the $S$ with the plane passing through the points $Z$ and $W$ such that the intersection is great circle, say $C$. Suppose that chord joining point $Z$ and $W$ makes angle $\theta$ to center (origin $O$) of $C$. During this process we will get  triangle $OZW$ lies in that plane, where $OZ = 1$ and $OW = 1$ and we know that Euclidean distance between $Z$ and $W$ is $d_c(z,w)$.
Now using the Arc Length Formula of circle we have,
\begin{align}
     \theta = d_s(z,w)
\end{align}
Also we know that if we draw angle bisector of $\theta$, say $OM$,we get (N0te that $OZW$ is isosceles triangle),
\begin{align}
    2Sin(\frac{\theta}{2}) = d_c(z,w) 
\end{align}
We have
\begin{align}
     2Sin(\frac{d_s(z,w)}{2}) = d_c(z,w)
\end{align}
Now using the simple trigonometric fact $\frac{2\theta}{\pi} \leq Sin\theta \leq \theta$, We have
\begin{align}
  \frac{2}{\pi} d_s(z,w) \leq d_c(z,w) \leq d_s(z,w)
\end{align}
Hence from this we proved that spherical metric $d_s$ and chordal metric $d_c$ are equivalent. If you want to get the same inequality as mentioned in the question you just need to play around the above trigonometric fact and above metric equivalent inequality.
