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Math Help - Complex analysis Inequality

  1. #1
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    Complex analysis Inequality

    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?
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  2. #2
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    Quote Originally Posted by problem View Post
    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.
    Last edited by Opalg; April 13th 2011 at 12:21 PM.
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