How do I prove that the metric in $\mathbb{C}_{\infty}$ satisfies the triangle inequality? To be specific, there is a logical way to prove that this metric satisfies the triangle inequality without using any magical identity ? 
$
\displaystyle{d(z,z') =
\begin{cases}
\displaystyle\frac{2|z - z'|}{[(1 + |z|^2)^(1 + |z'|^2)]^{\frac{1}{2}}}, & \mbox{ if } z, z' \in \mathbb{C} \\
\displaystyle\frac{2}{(1 + |z|^2)}, &  \mbox{ if } z \in \mathbb{C} \mbox{ and } z' = \infty
\end{cases}
}
$
Thanks!
 A: Just for reference, I completed the answer to my question provided by Alfonso Delfin 
$d(z,z') \leq d(z,z'') + d(z'',z')$
Let $\sigma: \mathbb{C}\cup\{\infty\} \to S$ be the stereographic projection
\begin{cases}
(\displaystyle\frac{2\Re(z)}{|z|^2+1},\displaystyle\frac{2\Im(z)}{|z|^2+1},\displaystyle\frac{|z|^2 - 1}{|z|^2 + 1}), & \mbox{ if } z \in \mathbb{C}\\
(0,0,1), & \mbox{ if } z=\infty
\end{cases}
Notice that
$d(z,z')=||\sigma(z)-\sigma(z')||$,
where $||\cdot||$ is the usual norm in $\mathbb{R}^3$. Thus, the triangle inequality follows from the one for $||\cdot||.$
Let $\textbf{u}, \textbf{v} \in \mathbb{R}^n$, then
$||\textbf{u} + \textbf{v}||^2 = (\textbf{u} + \textbf{v})\cdot(\textbf{u} + \textbf{v}) = ||\textbf{u}||^2 + 2(\textbf{u}\cdot\textbf{v}) + ||\textbf{v}||^2$
By Cauchy-Schwartz inequality we have
$
\textbf{u}\cdot\textbf{v} \leq ||\textbf{u}|| \mbox{ } ||\textbf{v}||
$
Hence,
$||\textbf{u} + \textbf{v}||^2 \leq ||\textbf{u}||^2 + 2||\textbf{u}|| \mbox{ } ||\textbf{v}|| + ||\textbf{v}||^2 = (||\textbf{u}|| + ||\textbf{v}||)^2
$
Taking square roots both sides yields
$||\textbf{u} + \textbf{v}|| \leq ||\textbf{u}|| + ||\textbf{v}||$
Now consider
$ d(z,z')= ||[\sigma(z)- \sigma(z'')] + [\sigma(z'') - \sigma(z')]||$
Let $\textbf{u}= \sigma(z) - \sigma(z''), \textbf{v} = \sigma(z'') - \sigma(z')$, then
$d(z,z')= ||\textbf{u} + \textbf{v}|| \leq ||\textbf{u}|| + ||\textbf{v}|| = d(z,z'') + d(z'',z') $
