Complex analysis Inequality

**problem** · April 13th 2011, 07:10 AM

Show that if $|a|<1$ and $|b|<1$ , then

$\frac{|a|-|b|}{1-|ab|} \le \frac{|a+b|}{|1+ab|} \le \frac{|a|+|b|}{1+|ab|}$ .

I try to prove by squaring the terms but it is long. Do anyone has a shorter approach to the question?

**Opalg** · April 13th 2011, 10:53 AM

Originally Posted by problem

Show that if $|a|<1$ and $|b|<1$ , then

$\frac{|a|-|b|}{1-|ab|} \le \frac{|a+b|}{|1+ab|} \le \frac{|a|+|b|}{1+|ab|}$ .

I try to prove by squaring the terms but it is long. Do anyone has a shorter approach to the question?

The right-hand inequality is false.

For example, let $a = b = \alpha i$ , where $0<\alpha<1.$ Then $\frac{|a+b|}{|1+ab|} = \frac{2\alpha}{1-\alpha^2}$ and $\frac{|a|+|b|}{1+|ab|} = \frac{2\alpha}{1+\alpha^2}.$ It is clearly not true that $\frac{2\alpha}{1-\alpha^2}\leqslant \frac{2\alpha}{1+\alpha^2}.$

Edit. The left-hand inequality is also false, as you can check by putting $a=\alpha i$ and $b=-\beta i$ , with $0<\beta<\alpha<1.$

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