TITLE: Show that $\left|\frac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1$ when $|\alpha|,|\beta| < 1$
QUESTION [16 upvotes]: This is the question I'm stumbling with:

When $|\alpha| < 1$ and $|\beta| < 1$, show that:
$$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1$$

The chapter that contains this question contains (among others) the triangle inequalities: 
$$\left||z_1| - |z_2|\right| \le |z_1 + z_2| \le |z_1| + |z_2| $$
I've tried to use the triangle inequalities to increase the dividend and/or decrease the divisor:
$$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < \cfrac{|\alpha| +|\beta|}{\left|1-|\bar{\alpha}\beta|\right|}$$
But it's not clear if or why that would be smaller than one.  I've also tried to multiply the equation by the conjugated divisor $\cfrac{1-\alpha\bar{\beta}}{1-\alpha\bar{\beta}}$, which gives a real divisor, but the equation does not appear solvable.
Any hint would be much appreciated.

REPLY [2 votes]: A silly answer, given the context - but:
Consider the function
$$f(z)= {z -\beta\over 1- z\overline \beta},$$
where $|\beta|<1$, with domain the disc $|z| \le 1$.
The denominator does not vanish: if $\beta\not = 0$,
$$ 1-z\overline\beta = 0 \iff z = \beta/|\beta|^2,$$
requiring $|z|>1$.
Hence $f$ is analytic on the disc.
Now, if $|z|=1$, then $z^{-1}= \overline z$, so that
$$|f(z)|= \left|{z -\beta\over 1- z\overline \beta}\right| = \left|{z -\beta\over \overline z- \overline \beta}\right| = 1.$$
So, by the maximum modulus principle, $|f(\alpha)|< 1$, if $|\alpha| <1.$
As I recall, I saw this argument in Rudin's RCA.