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Math Help - An inequality

  1. #1
    Senior Member bkarpuz's Avatar
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    Cool An inequality

    Prove that |\mathrm{Re}(z)|+|\mathrm{Im}(z)|\leq\sqrt{2}\|z\| for all z\in\mathbb{C}.

    Not very hard, but I wonder to see if there will be nice solutions.
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  2. #2
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    Quote Originally Posted by bkarpuz View Post
    Prove that |\mathrm{Re}(z)|+|\mathrm{Im}(z)|\leq\sqrt{2}\|z\| for all z\in\mathbb{C}.

    Not very hard, but I wonder to see if there will be nice solutions.
    a^2 + b^2 \geq 2|a||b| is equivalent to (|a|+|b|)^2 \leq 2(a^2+b^2).
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  3. #3
    Senior Member bkarpuz's Avatar
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    Red face

    Quote Originally Posted by NonCommAlg View Post
    a^2 + b^2 \geq 2|a||b| is equivalent to (|a|+|b|)^2 \leq 2(a^2+b^2).
    Of course this is the most simple proof.

    And still waiting for interesting solutions.
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  4. #4
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    If z=r(\cos\theta+i\sin\theta)\mbox{ with } r=|z| then

    |\mathrm{Re}(z)|+|\mathrm{Im}(z)|=r(|\cos\theta|+|  \sin\theta|)=r\sqrt{1+|\sin 2\theta|}\leq r\surd2.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by halbard View Post
    If z=r(\cos\theta+i\sin\theta)\mbox{ with } r=|z| then

    |\mathrm{Re}(z)|+|\mathrm{Im}(z)|=r(|\cos\theta|+|  \sin\theta|)=r\sqrt{1+|\sin 2\theta|}\leq r\surd2.
    My own solution was also similar.
    Just set f(t):=|t|+|\sqrt{1-t^{2}}| for t\in[-1,1], and show that f(t)\leq f(\sqrt{2}/2)=\sqrt{2} for all t\in[0,1].
    Since f is even, we have f(\cos(t))\leq\sqrt{2} for all t\in[0,2\pi].
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