Prove that $\left|\frac{z_1-z_2}{1-z_1\bar{z_2}}\right|\lt 1$ Question: Prove that $\left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|\lt 1$ if $|z_1|\lt1$, $ |z_2|\lt 1$

My solution:
I had no idea how to go about this one so instead I started simplifying the inequality and my solution is as follows:-
$$\begin{equation}
\left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|\lt 1 \\
\implies \left|\dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right|^2\lt 1 \\ 
\implies \left( \dfrac{z_1-z_2}{1-z_1\bar{z_2}}\right)\overline{\left({\dfrac{z_1-z_2}{1-z_1\bar{z_2}}}\right)} \lt 1 \\
\implies |z_1|^2+|z_2|^2 \lt 1 +|z_1|^2|z_2|^2 \\
\implies (|z_1|^2-1)(1-|z_2|^2) \lt 0
\end{equation}$$
Now as $|z_1| \lt 1$, so $(|z_1|^2-1) \lt 0$ and similarly $(1-|z_2|^2) \gt 0$, hence we can safely conclude that the above inequality holds.

Whats my question about:-
Now the book that I am solving from also gave the same solution it just further factored $(|z_1|^2-1)(1-|z_2|^2)$. This doesn't seem a good enough proof for me. If anyone can suggest a proof which doesn't simply verify the statement to be proved.

P.S.:- This question has also been asked here, but I don't think I have the same query as the OP of that post.
 A: As has been pointed out, this only solves the problem when we maximise the numerator, which of course doesn't give the full answer. At the moment I cannot see how to solve it in general, but I'll leave this answer just as a start. 
Maybe one can try by fixing $r_1,\phi_1,r_2$ and seeing the difference as a function of $\phi_2$ and try to minimize it. But it's not what I was looking for
Note that two points $z_1=r_1e^{i\phi_1}$ and $z_2=r_2e^{i\phi_2}$ have the maximal distance if $\phi_2=\phi_1+i\pi$, i.e. they are symmetric with respect to the center of the circle.
Their distance is then simply given by
$$|z_1-z_2|=r_1+r_2.$$
but now $$z_1\overline{z_2}=r_1r_2e^{i\pi}=-r_1r_2$$
so that
$$|1-z_1\overline{z_2}|=1+r_1r_2.$$
Now the proof follows by showing that for $r_i<1$ 
$$1+r_1r_2>r_1+r_2$$
Which is equivalent and follows from the fact that
$$(1-r_1)(1-r_2)>0.$$
A: For $\left|z\right|\le1$ and $-1\le x\le1$, we have
$$
\begin{align}
1-\left|\frac{(1+z)x}{1+zx^2}\right|^2
&=1-\frac{(1+z)x}{1+zx^2}\frac{(1+\bar z)x}{1+\bar zx^2}\\
&=\frac{\left(1-x^2\right)\left(1-\left|z\right|^2x^2\right)}{1+2\mathrm{Re}(z)x^2+\left|z\right|^2x^4}\\
&\ge\frac{\left(1-x^2\right)\left(1-\left|z\right|^2x^2\right)}{\left(1+\left|z\right|x^2\right)^2}\\[6pt]
&\ge0\tag{1}
\end{align}
$$
Assume, wlog, that $\left|z_1\right|\le\left|z_2\right|$. Then, setting $z=-\frac{z_1}{z_2}$ and $x=\left|z_2\right|$ in $(1)$, we get
$$
\begin{align}
\left|\frac{z_1-z_2}{1-z_1\bar{z}_2}\right|
&=\left|\frac{\left(1-\frac{z_1}{z_2}\right)\left|z_2\right|}{1-\frac{z_1}{z_2}\left|z_2\right|^2}\right|\\[6pt]
&\le1\tag{2}
\end{align}
$$
A: All denominators in your calculation are strictly positive, and multiplying
an inequality with a positive number gives an equivalent inequality.
Therefore, as already pointed out in the comments, the implications are
two-sided and you can simply replace $\Longrightarrow$ by
$\Longleftrightarrow$ to get a fully valid proof.
That is a  common procedure if you try to prove something: Start with
the claim and try to find a "simpler" equivalent statement. 
If that task was successful then you can try to find a "direct" proof.
In your case, a possible approach would be to calculate
$$
\lvert 1-z_1\bar{z_2} \rvert^2 - \lvert z_1-z_2 \rvert^2 = ... = (1 - \lvert z_1 \rvert^2)(1 - \lvert z_2 \rvert^2)
$$
(which is valid for all complex number $z_1, z_2$).
For $\lvert z_1 \rvert < 1, \lvert z_2 \rvert < 1$ the expression is positive, and 
$$
 \left\lvert \frac{z_1-z_2}{1-z_1\bar{z_2}}\right \rvert\lt 1
$$
follows. As a "bonus", you see immediately what happens if
both $z_1, z_2$ are outside of the unit circle, or one is inside 
and the other is outside, without repeating the calculation.
