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Math Help - inequality proof

  1. #1
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    inequality proof

    I'm a little stuck with the following inequality and was wondering if anyone could help me out.

    \frac{\mbox{Re}(z_1+z_2)}{|z_3+z_4|} \leq \frac{|z_1|+|z_2|}{||z_3|-|z_4||} \ \ \mbox{where} \ \ z_3 \neq z_4


    \frac{x_1+x_2}{|z_3+z_4|} \leq \frac{|z_1|+|z_2|}{||z_3|-|z_4||}


    \frac{x_1+x_2}{|z_3+z_4|} \leq \frac{\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}}{||z_3  |-|z_4||}

    at which point I get stuck.

    I was also thinking of using the property that:

    |z_3+z_4| \geq ||z_3|-|z_4|| \therefore \frac{1}{|z_3+z_4|} \leq \frac{1}{||z_3|-|z_4||} but the numerators are throwing me off.
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  2. #2
    Moo
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    Hello,

    I was also thinking of using the property that:

    |z_3+z_4| \geq ||z_3|-|z_4||
    What is this property ?

    but the numerators are throwing me off.
    By the numerators, were you thinking of x_1+x_2 \le \sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2} ?
    \sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2} \ge \sqrt{x_1^2}+\sqrt{x_2^2}=|x_1|+|x_2| \ge x_1+x_2
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  3. #3
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    For the numerator, use Moo's idea, and for the denominator, use indeed yours (the lower bound in the triangular inequality).

    About this lower bound, it is a consequence of the usual triangle inequality: for any x,y, |x|\leq|x+y|+|-y|, hence |x|-|y|\leq |x+y| and by symmetry the same holds swapping x and y, so that ||x|-|y||=\max(|x|-|y|,|y|-|x|)\leq |x+y|.

    Laurent.
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