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Let $a_i$ and $b_i$, $i=1,\ldots,n$, be two finite sequences of numbers in $(0,1)$. We know that $a_i < b_i$ for all $i$. Is it then true that $a_1\cdots a_n < b_1\cdots b_n$?

If all the $a_i$ were the same, and all the $b_i$ were the same, this would be rather obvious since the functions $x^n$ for positive $n$ are known to be increasing on $(0,1)$. So in this special case, it works. But I'm not sure how to generalize this result.

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up vote 5 down vote accepted

If $x>y>0$ and $u>v>0$, then we always have $xu>xv$ and $xv>yv$, so that $xu>yv$.

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$a_1\cdots a_n<a_1\cdots a_{n-1}b_n<\cdots < a_1\cdots a_{k-1}b_k\cdots b_n<\cdots <a_1b_2\cdots b_n<b_1\cdots b_n$.

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