Show that $\left|\frac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1$ when $|\alpha|,|\beta| < 1$ 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.
 A: The hint of mrf gives a nicest proof, of course, but we can use here also the following algebra:
Let $\alpha=x+yi$ and $\beta=a+bi,$ where $\{x,y,a,b\}\subset\mathbb R.$ 
Thus, we need to prove that
$$|x-a+(y-b)i|<|1-(x-yi)(a+bi)|$$ or
$$(x-a)^2+(y-b)^2<(1-ax-by)^2+(ay-bx)^2$$ or
$$x^2+y^2+a^2+b^2<1+a^2x^2+b^2y^2+a^2y^2+b^2x^2$$ or
$$1-(x^2+y^2+a^2+b^2)+(x^2+y^2)(a^2+b^2)>0$$ or
$$(1-x^2-y^2)(1-a^2-b^2)>0,$$ which is obvious.
A: 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.
A: Hint:
$$
\left| \frac{\alpha-\beta}{1-\bar\alpha\beta}\right| < 1 \Leftrightarrow |\alpha-\beta|^2 < |1-\bar\alpha\beta|^2.
$$
Expand both sides, remembering that $|z|^2 = z\bar z$ and simplify. That should get you where you want to be after some algebra.
A: Using $z \cdot \bar{z}=|z|^{2}$ yields
$$
\begin{aligned}
|1-\bar{\alpha} \beta|^{2}-|\alpha-\beta|^{2} &\left.=1+|\alpha|^{2}|\beta|^{2}-|\alpha|^{2}-\mid \beta\right)^{2} \\
&=\left(1-|\alpha|^{2}\right)\left(1-|\beta|^{2}\right) \\
&>0 \\
\therefore \quad |1-\bar{\alpha} \beta|^{2}&>|\alpha-\beta|^{2} \\
|1-\bar{\alpha} \beta|&>| \alpha-\beta \mid \end{aligned}
$$
Hence we may conclude that
$$\left|\frac{\alpha-\beta}{\mid-\bar{\alpha} \beta}\right|<1. $$
