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Proving an inequality

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data

    If z and w are complex numbers such that |z|<=1 and |w|<=1 then prove

    [tex]\left| z+w \right| \leq \left| 1 + \overline{z} w \right| [/tex]

    3. The attempt at a solution

    I have reduced this to essentially

    x^2+y^2 <= 1+(xy)^2.

    It seems to me if both x and y are less than or equal to 1, then the inequality must hold. I can't think of how to prove this formally, though. Any help on how to do this would be appreciated.
     
  2. jcsd
  3. Jan 21, 2010 #2

    Dick

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    Ok, so if x^2=a and y^2=b then you want to prove a+b<=1+ab, where 0<=a<=1 and 0<=b<=1. Write it as a*(1-b)+1*b<=1. If b is in [0,1] then the left side is a number between 'a' and '1', right? There's a name for this kind of inequality, but I forget what it is. There may also be an easier way to prove this. But I forget that too.
     
  4. Jan 21, 2010 #3

    Dick

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    Maybe it's a form of a Jensen's inequality, in kind of a vague way.
     
    Last edited: Jan 21, 2010
  5. Jan 21, 2010 #4
    Dick,

    Thanks for your post. I am sorry, but I do not see how/why the following is true:

     
  6. Jan 22, 2010 #5
    Suppose [tex] |\bar{z}| < 1 [/tex] then


    [tex]
    |w|^2 \le \frac{1-|z|^2}{1-|z|^2}
    [/tex]


    [tex]
    |w|^2-|\overline{z} w|^2 \le 1-|z|^2
    [/tex]


    [tex]
    |w|^2 + |z|^2 + 2\mbox{ Re}(\overline{z} w) \le 1 + |\overline{z} w|^2 + 2\mbox{ Re}(\overline{z} w)
    [/tex]


    [tex]
    \left| z+w \right|^2 \leq \left| 1 + \overline{z} w \right|^2
    [/tex]
     
    Last edited: Jan 22, 2010
  7. Jan 22, 2010 #6

    Dick

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    Can you show if t is in [0,1] then f(t)=a*t+b*(1-t) is between a and b? f(0)=b, f(1)=a and f(t) is a linear function.
     
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