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 $$\left|{\frac{z_1-z_2}{1-z_1 \overline{z_2}}}\right| < 1$$
?
