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Thread: Complex analysis Inequality

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by problem View Post
    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 $\displaystyle 0<\beta<\alpha<1.$
    Last edited by Opalg; Apr 13th 2011 at 11:21 AM.
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