# Complex analysis Inequality

• Apr 13th 2011, 07:10 AM
problem
Complex analysis Inequality
Show that if $\displaystyle |a|<1$ and $\displaystyle |b|<1$ , then

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

I try to prove by squaring the terms but it is long. Do anyone has a shorter approach to the question?
• Apr 13th 2011, 10:53 AM
Opalg
Quote:

Originally Posted by problem
Show that if $\displaystyle |a|<1$ and $\displaystyle |b|<1$ , then

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

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

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