问题征解[27]代数不等式(已解决)

已知$x_1,x_2,\cdots ,x_n>0$,求证:$$\dfrac{1}{x_1}+\dfrac{2}{x_1+x_2}+\cdots +\dfrac{n}{x_1+x_2+\cdots +x_n}<4\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots +\dfrac{1}{x_n}\right).$$

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问题征解[27]代数不等式(已解决)》有3条回应

  1. real0371说:

    lanqi老师v5!!

  2. real0371说:

    let a_i=\frac{1}{x_i}
    \frac{n}{\Sigma_{i=1}^{n}x_i}=\frac{n}{\Sigma_{i=1}^{n}\frac{1}{a_i}\leq
    (\Pi_{i=1}^{n}a_{i})^{\frac{1}{n}}
    by Carlman inequality
    \Sigma_{k=1}^{n} (\Pi_{i=1}^{k}a_{i})^{\frac{1}{k}} \leq e\Sigma_{k=1}^{n} a_{k}
    <4\Sigma_{k=1}^{n} a_{k}=4\Sigma_{k=1}^{n}\frac{1}{x_k}

    • 意琦行说:

      这是一个已经解决的问题.附上解析.
      根据柯西不等式,有\[\dfrac {1^2}{x_1}+\dfrac {2^2}{x_2}+\cdots+\dfrac {n^2}{x_n}\geqslant \dfrac{(1+2+\cdots+n)^2}{x_1+x_2+\cdots+x_n}=\dfrac {n^2(n+1)^2}4\cdot \dfrac{1}{x_1+x_2+\cdots+x_n},\]于是\[\dfrac{n}{x_1+x_2+\cdots+x_n}\leqslant \dfrac{4}{n(n+1)^2}\cdot \left(\dfrac {1^2}{x_1}+\dfrac{2^2}{x_2}+\cdots+\dfrac{n^2}{x_n}\right), \]于是\[\begin{split} LHS&=\sum_{k=1}^n\dfrac{k}{x_1+x_2+\cdots+x_k}\\ &\leqslant \sum_{k=1}^n\dfrac{4}{k(k+1)^2}\left(\dfrac 1{x_1}+\dfrac{2^2}{x_2}+\cdots+\dfrac{k^2}{x_k}\right)\\ &=\sum_{k=1}^n\dfrac{4k^2}{x_k}\left(\sum_{i=k}^n\dfrac{1}{i(i+1)^2}\right)\\ &<\sum_{k=1}^n\dfrac{4k^2}{x_k}\sum_{i=k}^n\left(\dfrac{1}{2i^2}-\dfrac{1}{2(i+1)^2}\right)\\ &=\sum_{k=1}^n\dfrac{4k^2}{x_k}\cdot\left(\dfrac{1}{2k^2}-\dfrac{1}{2(n+1)^2}\right)\\ &<2\sum_{k=1}^n\dfrac{1}{x_k}\\ &=RHS,\end{split}\]因此原命题得证.

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