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Math Help - Proof Q

  1. #1
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    Proof Q

    Show that \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}

    I was thinking along the lines of dividing both sides by \sqrt[3]{3}, rearranging and comparing the cubes of both sides but doesnt seem to work.
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  2. #2
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    Quote Originally Posted by vuze88 View Post
    Show that \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}

    I was thinking along the lines of dividing both sides by \sqrt[3]{3}, rearranging and comparing the cubes of both sides but doesnt seem to work.
    There is probably a simpler proof, but here is what i suggest:

    We go to a proof by contradiction. Assume that is valid:

    <br />
\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}} > 2\sqrt[3]{3}

    We raise both sides to cube:
    <br />
(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}})^3 > (2\sqrt[3]{3})^3

    By binomial expansion we have:
    <br />
6+3(3-3^{1/3})^{2/3}(3+3^{1/3})^{1/3}+3(3-3^{1/3})^{1/3}(3+3^{1/3})^{2/3} > 24

    By extracting as common factor the term:
    (3^2-3^{2/3})^{1/3} we do have:

    <br />
(3^2-3^{2/3})^{1/3}(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}})> 18

    But we know that \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}} > 2\sqrt[3]{3}

    So by substituting we have:
     <br />
(3^2-3^{2/3})^{1/3}2\sqrt[3]{3} > 18<br />
    Or
     <br />
6(3^3-3\sqrt[3]{3})^{1/3} > 18\Rightarrow (3^3-3\sqrt[3]{3})^{1/3} > 3<br />
    By raising again to cube we have that:
     <br />
3^3-3\sqrt[3]{3} > 3^3\Rightarrow-3\sqrt[3]{3} > 0<br />
    which is false.
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  3. #3
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    Quote Originally Posted by vuze88 View Post
    Show that \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}

    I was thinking along the lines of dividing both sides by \sqrt[3]{3}, rearranging and comparing the cubes of both sides but doesnt seem to work.
    If you know what that means, you can use the strict concavity of the cubic root function. Id est: \sqrt[3]{\frac{x+y}{2}}>\frac{\sqrt[3]{x}+\sqrt[3]{y}}{2} when 0\leq x\neq y. With x,y=3\pm\sqrt[3]{3}. This is the "visual" proof (you can see the inequality on a plot of \sqrt[3]{}) and it immediately generalizes to \sqrt[3]{a+b}+\sqrt[3]{a-b}< 2\sqrt[3]{a} for any 0<b<a.
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