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Thread: Inequality cubic-radical

  1. #1
    Super Member dhiab's Avatar
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    Inequality cubic-radical

    Hello : is the positifs reales numbers :
    prove :
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  2. #2
    Senior Member pankaj's Avatar
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    What you probably mean is that if 'a' and 'b' are positive real numbers then prove that b\geq 3\sqrt[3]{ab}-2\sqrt{a}.

    Solution

    Let x=\sqrt{a}.Let y=b

    Since x and y are positive real numbers,therefore

    \frac{x+x+y}{3}\geq \sqrt[3]{x^2y} since A.M > G.M

    2\sqrt{a}+b\geq 3\sqrt[3]{ab}

    b\geq 3\sqrt[3]{ab}-2\sqrt{a}.
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