Can't prove statement jumped over in book proof, I actually think it might be wrong. But it is used throughout the book I wish to prove that the metric $d(x,y)=\sum^\infty_{n=1}\frac{1}{2^n}\frac{|x_n-y_i|}{1+|x_n-y_n|}$
This is a metric on infinite sequences, x,y and z are sequences with terms $x_i$ and so forth.
The problem is the triangle inequality requirement, $d(x,y)\le d(x,z)+d(z,y)$
$\sum^\infty_{n=1}\frac{1}{2^n}\frac{|x_n-y_i|}{1+|x_n-y_n|}=
\sum^\infty_{n=1}\frac{1}{2^n}\frac{|x_n-z_n+z_n-y_i|}{1+|x_n-z_n+z_n-y_n|}$
$\le\sum^\infty_{n=1}\frac{1}{2^n}\frac{|x_n-z_i|}{1+|x_n-z_n+z_n-y_n|}+\sum^\infty_{n=1}\frac{1}{2^n}\frac{|z_n-y_i|}{1+|x_n-z_n+z_n-y_n|}$
Now I want to show something like $\frac{1}{1+|x_n-z_n+z_n-y_n|}=\frac{1}{1+|x_n-y_n|}\le\frac{1}{1+|x_n-z_n|}$
$\iff1+|x_n-y_n|\ge1+|x_n-z_n|$
So, I've been playing with this for a while and have got nowhere. The term with coefficient $\frac{1}{2^n}$ is itself a metric, so it should be true (the series part isn't important).
But why can't $x=3,y=2,z=1$ be valid, then $1+|3-2|=2\ge1+|3-1|=3$ which is false! Even if you change the inequality sign (so the other way instead) just set y=1 and z=2 and the result is contradicted!

Note that $\frac{d}{da}\frac{a}{1+a} = \frac{1}{(a+1)^2} > 0$, so $\frac{a}{1+a} \leq \frac{b}{1+b}$ if $b \geq a$. Let $$b := d(x,z) + d(z,y),\quad a:= d(x,y).$$
By the above, we have
\begin{align*}
\frac{d(x,y)}{1+d(x,y)} &\leq \frac{d(x,z) + d(z,y)}{1+d(x,z)+d(y,z)}\\
&= \frac{d(x,z)}{1+d(x,z) + d(y,z)} + \frac{d(y,z)}{1+d(y,z)+d(x,z)} \\
&\leq \frac{d(x,z)}{1+d(x,z)} + \frac{d(y,z)}{1+d(y,z)}.
\end{align*}
The last inequality follows because $d(\cdot,\cdot) \geq 0.$
