Simple Inequality of Complex Numbers, $\left| \frac{a-b}{1-\overline{a}b} \right| <1$ Exercise from Ahlfor's Complex:
Given $a,b \in \mathbb C$, with $|a| <1$, $|b|<1$, prove:
$$\left|\frac{a-b}{1-\overline{a}b}\right| <1.$$  
My argument:
Lemma:   
If $\alpha, \beta \in \mathbb R$ and $\alpha, \beta <1$, then $\alpha^2+\beta^2 < 1 + \alpha^2\beta^2$.  
To prove the lemma, just write $\alpha = (1-\epsilon_1)$, $\beta = (1-\epsilon_2)$ with $0 < \epsilon_i < 1$. Then, expanding out both sides, we get
$$ 1+\alpha^2\beta^2 = \alpha^2 + \beta^2 + 4\epsilon_1\epsilon_2 - 2\epsilon_1^2\epsilon_2 - 2\epsilon_1\epsilon_2^2 + \epsilon_1^2\epsilon_2^2,$$ but $\epsilon_i^2 < \epsilon_i$, so $$4\epsilon_1\epsilon_2 - 2\epsilon_1^2\epsilon_2 - 2\epsilon_1\epsilon_2^2 > 0.$$
The result follows.  
Now, $$\begin{eqnarray*}
\left|\frac{a-b}{1-\overline{a}b}\right|  &=& \frac{|a-b|}{|1-\overline{a}b|} \\
&\leq& \frac{|a-b|}{|1-|a||b| \, |} \\
\end{eqnarray*}
$$
$$
\begin{eqnarray*}
\frac{|a-b|^2}{|1-|a||b| \, |^2} 
&\leq& \frac{|a|^2+|b|^2-2|a||b|}{1+|a|^2|b|^2-2|a||b|} \\
&<& 1
\end{eqnarray*}$$
by the lemma, and since $x^2<1$ implies $x<1$ for $x \in \mathbb R$, the result follows.
Where I've used the reverse triangle inequality and the fact that $|z| \geq \Re(z).$  
Any slicker ways to do this? Any issues with this?
 A: We have 
$$
\begin{align*}
|1 - \overline{a}b|^2 - |a - b|^2 &= \left(1 - \overline{a}b \right)\left(1 - a\overline{b}\right) - (a - b)\left( \overline{a} - \overline{b}\right) \\
&= 1 - a\overline{a} - b\overline{b} + a\overline{a}b\overline{b} \\
&= \left( 1 - |a|^2\right)\left( 1 - |b|^2\right)\\
&> 0.
\end{align*}
$$
It follows that $|1 - \overline{a}b| > |a - b|$, hence
$$\left| \frac{a-b}{1 - \overline{a}b}\right| = \frac{|a-b|}{\left| 1 - \overline{a}b\right|} < 1.$$
A: A different approach can use the maximum modulus principle.
Let $\phi(z) = \frac{a-z}{1-\bar a z}$. This function is analytic except at $z_0=1/\bar a$ for which $|z_0| > 1$, since $|a|=|\bar a| < 1$.
Now suppose that $|z|=1$. That is $z = e^{i\theta}$ for some $\theta \in \mathbb{R}$.
Then we have $$|\phi(e^{i\theta})| = \left| \frac{a-e^{i\theta}}{1-\bar a e^{i\theta}} \right| = \left| e^{-i\theta} \frac{a-e^{i\theta}}{e^{-i\theta}-\bar a }\right| = | e^{-i\theta} | \left| \frac{a-e^{i\theta}}{\overline{a-e^{i\theta}}} \right| = 1.$$
Since $\phi$ is not constant, and $\phi$ is analytic in the disc of radius $1$, we have (by the maximum modulus principle) $$|\phi(z)| < 1 \text{ for } |z| < 1.$$
