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Math Help - Use induction to prove the AM-GM inequality

  1. #1
    MHF Contributor alexmahone's Avatar
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    Use induction to prove the AM-GM inequality

    My attempt:

    For n=1, we have a_1=a_1, which is trivially true.
    Let the statement be true for some n.

    \frac{a_1+a_2+...+a_n}{n}\geq(a_1a_2...a_n)^{1/n}

    \Leftrightarrow a_1+a_2+...+a_n\geq n(a_1a_2...a_n)^{1/n}

    We need to prove that a_1+a_2+...+a_{n+1}\geq(n+1)(a_1a_2...a_{n+1})^{1/(n+1)}

    We know that a_1+a_2+...+a_{n+1}\geq n(a_1a_2...a_n)^{1/n}+a_{n+1}

    How do I show that n(a_1a_2...a_n)^{1/n}+a_{n+1}\geq (n+1)(a_1a_2...a_{n+1})^{1/(n+1)}?
    Last edited by alexmahone; August 11th 2011 at 09:06 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Use induction to prove the AM-GM inequality

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