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    \prod_{i=1}^n p_{i}^{p_i}\leq(\sum_{i=1}^n \sqrt{p_i})^2

    where the p_i sum to one.
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    it seems impenetrable.
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    it seems impenetrable.
    Maybe try this (im totally guessing here btw!)

     ln[\prod_{i=1}^n p_{i}^{p_i}]\leq ln[(\sum_{i=1}^n \sqrt{p_i})^2]

    using log rules LHS becomes

     \sum_{i=1}^n p_{i}ln(p_i) = \sum_{i=1}^n p_{i}\times \sum_{i=1}^n ln(p_i) =   \sum_{i=1}^n ln({p_i})

    as

     \sum_{i=1}^n p_i = 1

    from question.
    Then take ln of RHS and hopefully this can be proved to be greater than LHS?

    EDIT

    sorry this isn't even right, but maybe the method will still work?

    FURTHER EDIT

    No this method fails. Sorry - im out!
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    First thoughts..

    Take \phi = \log x

    Then by Jensen, we obtain:

    \displaystyle \sum p_i \log \sqrt{p_i} \leq \log \left ( \frac{\sum p_i \sqrt{p_i}}{\sum p_i} \right ) = \log \left ( \sum p_i \sqrt{p_i} \right )

    But p_i \sqrt{p_i} \leq \sqrt{p_i} from which the result follows.
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    More thoughts...

    The inequality "feels" quite weak. I mean, for p_i = 1/n the equation is:

    1/n <= n

    Weighted AM-GM gives us that we just need to prove

    \displaystyle \sum p_i^2 \leq (\sum \sqrt{p_i} )^2 which is also "easy" when expand and note that p_i^2 \leq p_i
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    ah, i see. thanks.
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    is this degree maths?
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    (Original post by Jfranny)
    is this degree maths?
    Probably, Chewwy is a third year at Cambridge
 
 
 
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