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consider this matrix $$\begin{bmatrix} a_{1,1}&a_{1,2}&\cdots,&a_{1,n}\\ a_{21}&a_{2,2}&\cdots&a_{2,n}\\ \cdots&\cdots&\cdots&\cdots\\ a_{m,1}&a_{m,2}&\cdots&a_{m,n} \end{bmatrix}$$ where $a_{i,j}>0$,

and define $$S^{(k)}_{i}=\sum_{1\le j_{1}<j_{2}<\cdots<j_{k}\le n}\prod_{v=1}^{k}a_{i,j_{v}}$$ $$S^{(0)}_{i}=1,i,k\in\{1,2,3,\cdots,n\}$$

show that $$\prod_{k=1}^{n}\left(\sum_{i=1}^{m}\dfrac{S^{(k)}_{i}}{S^{(k-1)}_{i}}\right)\le\prod_{i=1}^{n}\left(\sum_{k=1}^{m}a_{k,i}\right)$$

This inequality How prove it,Thank you

math110
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    Where does it show up? – Alt Feb 06 '14 at 20:21
  • It's china student creat, – math110 Feb 08 '14 at 16:28
  • I am not sure but here comes my suggestion: Since all terms are positive, it seems to me, it might be possible to show the inequality for each single factor (i.e. fixed $k$). Moreover, it might help to express $S_i^{(k)}$ as a sum over $\sigma \in PR(n,k)$ where $PR(n,k)$ stands for "permutations of $k$ elements among $n$ with repetition". I mean: $$S_i^{(k)} = \sum_{\sigma \in PR(n,k)} a_{i,\sigma(1)}\cdots a_{i,\sigma(k)}$$ – Martingalo Feb 09 '14 at 21:36

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