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

Question

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

Answer 1

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.