In Nehari's book Conformal Mapping he gives it as an exercise to prove that for $a,b\in \mathbb{C}$, $|a|, |b| <1$ we have $$\frac{|a|-|b|}{1-|ab|} \leq \left|\frac{a-b}{1-\overline{a}b}\right| \leq \frac{|a|+|b|}{1+|ab|}.$$I've been trying to prove this inequality for a while without success. Obviously the trouble is that the triangle inequality is making both numerator and denominator smaller (larger) on the left (right), making the inequality inconclusive at first glance. Things I have tried include:

  1. Multiplying each expression by one of the denominators to get rid of the fractions
  2. Exploiting the mapping $z\mapsto \frac{z-a}{1-\overline{a}z}$, which preserves the unit disk and takes $a\mapsto 0$.

So far no success. Does someone have some advice?


Why not straightforward approach? E.g. \begin{align} \left|\frac{a-b}{1-\bar{a}b}\right|^2-\left(\frac{|a|-|b|}{1-|ab|}\right)^2&= \frac{|a|^2+|b|^2-(\bar{a}b+a\bar{b})}{1+|ab|^2-(\bar{a}b+a\bar{b})}-\frac{|a|^2+|b|^2-2|ab|}{1+|ab|^2-2|ab|}=\\ &=\frac{(1-|a|^2)(1-|b|^2)(2|ab|-\bar{a}b-a\bar{b})}{|1-\bar{a}b|^2(1-|ab|)^2}\geq0, \end{align} since $|a|,|b|<1$ and $|ab|\geq \mathrm{Re}(\bar{a}b)$.

Similarly, \begin{align} \left(\frac{|a|+|b|}{1+|ab|}\right)^2-\left|\frac{a-b}{1-\bar{a}b}\right|^2&=\frac{|a|^2+|b|^2+2|ab|}{1+|ab|^2+2|ab|}- \frac{|a|^2+|b|^2-(\bar{a}b+a\bar{b})}{1+|ab|^2-(\bar{a}b+a\bar{b})}=\\ &=\frac{(1-|a|^2)(1-|b|^2)(2|ab|+\bar{a}b+a\bar{b})}{|1-\bar{a}b|^2(1+|ab|)^2}\geq0. \end{align}

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  • $\begingroup$ Thank you! I think squaring it is the key $\endgroup$ – Eric Auld Jun 17 '13 at 23:19

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