TITLE: Show $\left|{\frac{z_1-z_2}{1-z_1 \overline{z_2}}}\right| < 1$ if $|z_1| ,|z_2| < 1$
QUESTION [3 upvotes]: Show $$\left|{\frac{z_1-z_2}{1-z_1 \overline{z_2}}}\right| < 1$$ if $|z_1| <1$ and $|z_2| < 1$

Consider: $$\left|{\frac{z_1-z_2}{1-z_1 \overline{z_2}}}\right|^2$$
$$={\frac{|z_1-z_2|^2}{|1-z_1 \overline{z_2}|^2}}$$
$$=\frac{(z_1-z_2)(\overline{z_1-z_2})}{{(1-z_1\overline{z_2})(\overline{1-z_1\overline{z_2}})}}$$
$$=\frac{(z_1-z_2)(\overline{z_1}-\overline{z_2})}{{(1-z_1\overline{z_2})(1-\overline{z_1}z_2)}}$$
$$=\frac{z_1\overline{z_2}+z_2\overline{z_2}-z_1\overline{z_2}-\overline{z_1}z_2}{1+z_1\overline{z_2}\overline{z_1}z_2-\overline{z_1}z_2-z_1\overline{z_2}}$$
$$=\frac{|z_1|^2+|z_2|^2-2Re(z_1\overline{z_2})}{1+|z_1|^2|z_2|^2-2Re(z_1\overline{z_2})}$$
Now, I need to show that: $$|z_1|^2+|z_2|^2 < 1+|z_1|^2|z_2|^2$$
So, $$1+\frac{|z_1|^2}{|z_2|^2} < \frac{1}{|z_2|^2}+|z_1|^2$$
$$\rightarrow 1-|z_1|^2 < \frac{1}{|z_2|^2} - \frac{|z_1|^2}{|z_2|^2}$$
$$\rightarrow 1-|z_1|^2 < \frac{1 -|z_1|^2}{|z_2|^2}$$
This shows $$|z_1|^2+|z_2|^2-2Re(z_1\overline{z_2}) < 1+|z_1|^2|z_2|^2-2Re(z_1\overline{z_2})$$
So, $$|{\frac{z_1-z_2}{1-z_1 \overline{z_2}}}| < 1$$
Is this proof correct?
However I am also curious as to the following "proof".
$$|z_1|^2+|z_2|^2 < 1+|z_1|^2|z_2|^2$$
$$\rightarrow |z_1|^2 - |z_1|^2|z_2|^2 < 1-|z_2|^2$$
$$\rightarrow |z_1|^2(1-|z_2|^2) < 1-|z_2|^2$$
$$\rightarrow |z_1|^2< 1$$
I am curious because since $|z_1|<1$, then this implies $|z_1|^2 < 1$. So, is it sufficient to show that  $|z_1|^2 < 1$ which implies $$|z_1|^2+|z_2|^2 < 1+|z_1|^2|z_2|^2$$ which implies $$|z_1|^2+|z_2|^2-2Re(z_1\overline{z_2}) < 1+|z_1|^2|z_2|^2-2Re(z_1\overline{z_2})$$ which implies $$\left|{\frac{z_1-z_2}{1-z_1 \overline{z_2}}}\right| < 1$$
?

REPLY [4 votes]: Your calculations are correct, what you obtained is
$$|\,1- z_1 \bar z_2\,|^2 - |\,z_1-z_2\,|^2 =  1 + |\,z_1|^2 |\,z_2|^2- |\,z_1|^2 -|\,z_2|^2 $$
and the right hand side equals 
$$(1-|\,z_1|^2)\, (1-|\,z_2|^2) > 0$$ 
so you got it done.
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