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

    Solution

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

    Since $\displaystyle x$ and $\displaystyle y$ are positive real numbers,therefore

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

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

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