How prove The triangle inequality $\rho{(a,b)}\le \rho{(a,c)}+\rho{(c,b)}$ Question:

let $a,b,c$ be complex numbers,and such
  $$|a|<1,|b|<1,|c|<1$$
let $$\rho{(x,y)}=\left|\dfrac{x-y}{1-\overline{x}y}\right|$$
show that 
  $$\rho{(a,b)}\le \rho{(a,c)}+\rho{(c,b)}$$

we only prove
$$\left|\dfrac{a-b}{1-\overline{a}b}\right|\le \left|\dfrac{a-c}{1-\overline{a}c}\right|+\left|\dfrac{c-b}{1-\overline{c}b}\right|,|a|,|b|,|c|<1$$
then I can't ,
I  have found this book ,page 38,this book say this triangle inequality is less obvious,so I can't see anywhere have this inequality solution,
can you help me?
Thank you
 A: Reference is the book you referred :
Step 1 : Let $$ f(x,y)  :=\frac{x-y}{1-\overline{x} y } $$
From a routine computation we have $$ |f( f(x,y),f(x,z)) | = |f(y,z)
|$$
Since $$
 |f(a,b)| = |f(f(c,a),f(c,b))| = \bigg| \frac{ f(c,a) -
 f(c,b)}{1-\overline{f(c,a)} f(c,b)} \bigg| $$ we have a claim $$    \bigg| \frac{ f(c,a) -
 f(c,b)}{1-\overline{f(c,a)} f(c,b)} \bigg|\leq |  f(c,a) | +
 |
 f(c,b) |\ (1)$$
Step 2 : $$|x|,\ |y| < 1 \Rightarrow |f(x,y)| <1$$
Proof : It is followed from a direct computation.
Step 3 : That is if
$v:= f(c,a),\ w:= f(c,b) $ then we have
$$
 |v-w|\leq |v-|v|^2 w| +|w-|w|^2v |\ (2) \Leftrightarrow (1)$$
If $v:=v_1+iv_2,\ w:=w_1+iw_2$ then $$ \sqrt{(v_1-w_1)^2 +
 (v_2-w_2)^2} \leq \sqrt{(v_1-|v|^2w_1)^2 +
 (v_2-|v|^2w_2)^2} $$ $$+\sqrt{ (w_1-|w|^2v_1)^2 +
  (w_2-|w|^2v_2)^2 } \ (3)\Leftrightarrow (2) $$
Case 1 : $v_1w_1 \geq 0$ Then
$$ 0\leq |w|^4 v_1^2 + |v|^4 w_1^2- 2v_1w_1 (|v|^2+
 |w|^2-1)\Leftrightarrow  $$
$$0\leq (|w|^2 v_1 - |v|^2 w_1)^2 + 2v_1w_1 (1-|v|^2)(1- |w|^2)
$$
$$
\Rightarrow
  (v_1-w_1)^2 \leq (v_1-|v|^2w_1)^2 +
(w_1-|w|^2v_1)^2
$$
Case 2 : $v_1=-nw_1,\ n>0,\ v_2w_2 \geq 0$ Then $$
  (v_1-w_1)^2=(n+1)^2w_1^2, $$ $$  (v_1-|v|^2w_1)^2 +
(w_1-|w|^2v_1)^2  +2 |v_1-|v|^2w_1 || w_1-|w|^2v_1 | \geq $$ $$
(n+n^2w_1^2)^2w_1^2 +(1+nw_1^2)^2w_1^2 +
2(n+n^2w_1^2)(1+nw_1^2)w_1^2
$$
That is, we proved the following :
$$ (v_1-w_1)^2  \leq  (v_1-|v|^2w_1)^2  + (w_1-|w|^2v_1)^2 + 2
|v_1-|v|^2w_1 || w_1-|w|^2v_1 | $$
Case 3 : $v_1=-nw_1,\ v_2=-mw_2,\ n,\ m >0$
Note that
$$
nw_1^2+ mw_2^2 \leq \sqrt{n^2w_1^4 + (n^2+m^2)w_1^2 w_2^2 +
m^2w_2^4} $$
If we replace $ v_1=-nw_1,\ v_2=-mw_2$ in (3), then we have (3) from direct
computation. 
So we complete the proof.
