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Thread: Complex Inequalities

  1. #1
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    Complex Inequalities

    Suppose that $\displaystyle z$ and $\displaystyle w$ are in the closed unit disk.

    Prove the inequality $\displaystyle 1- \vert w \vert^2 \leq 2 \vert 1-z \bar w \vert$.

    I've tried to square both sides and expand the equation. However, I'm stuck there. Is there any other way that I can approach this question?

    Thanks in advance.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Hint :

    Use

    $\displaystyle |1-z\bar{w}|\geq |\;1-|z\bar{w}|\;|=1-|z||\bar{w}|\geq \ldots $


    Fernando Revilla
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  3. #3
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    $\displaystyle 2 \vert 1- z \bar w \vert
    \geq 2(\vert 1- \vert z \bar w \vert \vert )
    \geq 2(1- \vert zw \vert )
    \geq 2(1- \vert w \vert )$(Since $\displaystyle \vert z \vert \leq 1$)
    Now we claim that $\displaystyle 2 \vert w \vert - \vert w \vert^{2} \leq 1$.
    We let $\displaystyle y=2 \vert w \vert - \vert w \vert^{2}$ and substitute $\displaystyle \vert w \vert$ with $\displaystyle x$, where $\displaystyle 0 < x \leq 1$. Then we differentiate the equation $\displaystyle y$ to get the maximum point. From here we can get $\displaystyle y \leq 1$. Hence $\displaystyle 2 \vert w \vert - \vert w \vert^{2} \leq 1$.
    Therefore, $\displaystyle 2 \vert 1- z \bar w \vert
    \geq 2(1- \vert w \vert )
    \geq 2(1- \frac{1+ \vert w \vert^2}{2})
    =1- \vert w \vert^2$.

    Is there any easier method to this problem?
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