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Let $y_1,\ldots,y_{n+1}$ be positive real numbers satisfying $\displaystyle{\sum_{i=1}^{n+1} \frac{1}{ny_i+1}=1}$.

Is it true that $y_1y_2\cdots y_{n+1}\geq 1$?

Added: can we determine this inequality in terms of high-school math? (e.g. Cauchy-Schwarz inequality, arithmetic mean-geometric mean inequality)

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    Could you please consider using somewhat more descriptive titles? I'm just guessing you're the same person as the "Mark Jin" who asked [this question](http://math.stackexchange.com/questions/46900/on-the-inequality). Also, what do you want to achieve with these inequalities? No offense, I'm just curious.2011-06-23
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    I'm same person. I just want to prove $\sum_{i=1}^{n+1}\frac{1}{ny_i+1}=1$ implies $y_1\cdots y_{n+1}\geq 1$. From some Cauchy-Schwarz inequality, I can get $y_1+\cdots y_{n+1}\geq n+1$ and $1/y_1+1/y_2\cdots/y_{n+1}\geq n+1$. That's why I asked [that question](http://math.stackexchange.com/questions/46900/on-the-inequality) yesterday. And I'm unregistered user hence I should use another name.2011-06-23
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    @Jonas: I would prefer that `\displaystyle` was not used in question titles. It is still typeset as large inline text, and on the front page I feel it looks rather conspicuous.2011-06-23
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    @Rahul: I agree with that. Is this better, or still conspicuous?2011-06-23
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    @Theo: Looks good to me. Thanks!2011-06-23
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    @Rahul, Theo: I agree that it is much better now, thank you.2011-06-23
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    The cases $n=1,n=2$ are true.2011-06-23
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    @Mark: please register your account. You won't have any trouble with logging back in that way.2011-06-23

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