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Math Help - Inequality to prove involving convex notions

  1. #1
    Newbie Hugal's Avatar
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    Inequality to prove involving convex notions

    Hi everyone,

    Here's an exercice I have trouble with.
    Let a = (a_1,\dots,a_n)\in\mathbb{R}^n_+ and define G(a) = (a_1\dots a_n)^{\frac{1}{n}}

    I have to show that, assuming a,b\in\mathbb{R}^n_+, G(a+b) \geqslant G(a) + G(b).

    I have already tried a looooot of things. I think I'm suppose to use some convex inequalities : my first thought was to take the log of G(a+b), but it didn't seem to end very well for me.
    I've also tried to use other known inequalities, such as ab \leqslant \displaystyle \frac{a^2+b^2}{2}. But I don't think it works.

    So know, it's been almost an hour I'm on it, and I would like to have a small hint : I do not want the whole answer but just a tips which can make me progress.

    Thank you for your always-so-good answers,

    Hugo.
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  2. #2
    Super Member girdav's Avatar
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    Re: Inequality to prove involving convex notions

    We have to show that for 0\leq x_j \leq 1, we have \left(\prod_{j=1}^nx_j\right)^{\frac 1n}+\left(\prod_{j=1}^n(1-x_j)\right)^{\frac 1n}\leq 1. Take the n-th power, and use m:=\min_{1\leq j\leq n}x_j.
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