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Math Help - inequality proof

  1. #1
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    Smile inequality proof

    hello
    Let (x_1,x_2,...,x_n)\in \mathbb{R}^{*}_{+},by applying this inequality, \forall \alpha > 0,\log(\alpha )\leq \alpha -1 to every value of,
    \alpha _i=\frac{x_i}{\frac{x_1+x_2+...+x_n}{n}},i=1,2,..,  n.
    prove the following,
    \frac{1}{n}\sum_{k=1}^{n}\log x_k\leq \log(\frac{1}{n}\sum_{k=1}^{n}x_k)
    i believe it's the best place to post this question
    thanks.
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  2. #2
    Member Black's Avatar
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    For each i

    \text{log}(x_i)-\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\right  ) \le \frac{-x_1-\cdots+(n-1)x_i-\cdots-x_n}{x_1+\cdots+x_n}.

    As you combine the n inequalities, the right side will cancel out, leaving you with

    \sum_{k=1}^{n}\log(x_i)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t) \le 0.

    Divide by n to get the result.
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  3. #3
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    Smile

    would you please show me your steps,i don't get you.
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  4. #4
    Member Black's Avatar
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    We have the inequality \text{log}(\alpha) \le \alpha - 1 for nonnegative \alpha. So using \alpha_i we have the left side of the inequality:


    \text{log}(\alpha_i)=\text{log}\left(\frac{x_i}{\f  rac{x_1+\cdots+x_n}{n}}\right) = \text{log}(x_i)-\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\right  )

    and the right side:

    \alpha_i-1=\frac{x_i}{\frac{x_1+\cdots+x_n}{n}}-1=\frac{nx_i}{x_1+\cdots+x_n}-1=\frac{-x_1-\cdots+(n-1)x_i-\cdots-x_n}{x_1+\cdots+x_n}.

    There are n of these inequalities since i=1,2,\dots,n. We add the n inequalities together. For the left side, we will have

    \text{log}(x_1)+\cdots+\text{log}(x_n)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t)=\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t)

    and the right side:

    \frac{(n-1)x_1-\cdots-x_n}{x_1+\cdots+x_n}+\cdots+\frac{-x_1-\cdots+(n-1)x_n}{x_1+\cdots+x_n}=0,

    so

    <br />
\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t) \le 0

    Divide both sides by n to finish the proof.
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  5. #5
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    Smile

    Quote Originally Posted by Black View Post
    We have the inequality \text{log}(\alpha) \le \alpha - 1 for nonnegative \alpha. So using \alpha_i we have the left side of the inequality:


    \text{log}(\alpha_i)=\text{log}\left(\frac{x_i}{\f  rac{x_1+\cdots+x_n}{n}}\right) = \text{log}(x_i)-\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\right  )

    and the right side:

    \alpha_i-1=\frac{x_i}{\frac{x_1+\cdots+x_n}{n}}-1=\frac{nx_i}{x_1+\cdots+x_n}-1=\frac{-x_1-\cdots+(n-1)x_i-\cdots-x_n}{x_1+\cdots+x_n}.

    There are n of these inequalities since i=1,2,\dots,n. We add the n inequalities together. For the left side, we will have

    \text{log}(x_1)+\cdots+\text{log}(x_n)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t)=\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t)

    and the right side:

    \frac{(n-1)x_1-\cdots-x_n}{x_1+\cdots+x_n}+\cdots+\frac{-x_1-\cdots+(n-1)x_n}{x_1+\cdots+x_n}=0,

    so

    <br />
\sum_{k=1}^{n}\text{log}(x_k)-n\text{log}\left(\frac{1}{n}\sum_{k=1}^{n}x_k\righ  t) \le 0

    Divide both sides by n to finish the proof.
    thank you
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