Complex variables (prove Inequality)

Suppose that a, b∈C and that |a|＜1 and |b|＜1. Prove
that |(a-b)/(1-a(b‾))|＜1.
PS: C is complex number
b‾ is complex conjugates, if b=x+iy, b‾=x-iy

The inequality is equivalent to:

|a - b|^2 < |1 - ab‾|^2
<===> (a - b)‾(a - b) < (1 - ab‾)‾(1 - ab‾)
<===> a‾a - a‾b - b‾a + b‾b < 1 - ab‾ - a‾b + a‾ab‾b
<===> a‾a + b‾b < 1 + a‾ab‾b
<===> |a|^2 + |b|^2 < 1 + |a|^2|b|^2
<===> 0 < 1 - |a|^2 - |b|^2 + |a|^2|b|^2
<===> 0 < (1 - |a|^2)(1 - |b|^2)
<=== |a| < 1 and |b| < 1

which is true by hypothesis. So, the inequality holds. Moreover, it holds exactly when |a| < 1 and |b| < 1, or when |a| > 1 and |b| > 1. We would get equality if |a| = 1 or |b| = 1 (so long as the denominator is non-zero), and strict inequality will hold in the other direction if |a| > 1 and |b| < 1, or if |a| < 1 and |b| > 1.

b‾=x-iy