Which one is larger: $N\times\sum_{i=1}^N \frac{a_i}{b_i}$ or $\sum_{i=1}^N \frac{\sum_{i=1}^N a_i}{b_i}$? Here $N$ is a positive integer, $a_i>0$, $b_i>0$, $\forall i$. I only know that they are equal when $a_1=a_2=\ldots=a_N$ or $b_1=b_2=\ldots=b_N$.

  • $\begingroup$ I find some examples that $N \sum_{i=1}^N \frac{a_i}{b_i} \geq \sum_{i=1}^N \frac{\sum_{i=1}^N a_i}{b_i}$. Can we prove this result for arbitary $a_i$'s and $b_i$'s? $\endgroup$ – wbchu Aug 31 '16 at 11:46
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    $\begingroup$ if $a=(1,...,1,2)=b$, then $N\times\sum_{i=1}^N \frac{a_i}{b_i} = N^2$ and $\sum_{i=1}^N \frac{\sum_{i=1}^N a_i}{b_i} = (N+1)(N-\frac{1}{2})$ which is larger (when $N$ large). $\endgroup$ – anonymus Aug 31 '16 at 11:55
  • $\begingroup$ It seems that we can not draw a general result. Thanks. $\endgroup$ – wbchu Aug 31 '16 at 12:08

Nothing can be told about bigger number. Let us take $N=2$. Then we need to compare $$ 2\left(\frac{a_1}{b_1}+\frac{a_2}{b_2}\right) \quad \mathrm{vs.} \quad \left(a_1+a_2\right)\left(\frac{1}{b_1}+\frac{1}{b_2}\right) $$ which is equivalent to $$ \left(a_1-a_2\right)\left(\frac{1}{b_1}-\frac{1}{b_2}\right) \quad \mathrm{vs.} \quad 0. $$ And therefore any inequality is possible.

  • $\begingroup$ if $a_1 > a_2 > 0$ and $0<b_1<b_2$, then... etc $\endgroup$ – anonymus Aug 31 '16 at 12:08
  • $\begingroup$ @anonymus but there is no such a statement in the problem. If $a_i$ and $b_i$ are ordered in the way you state, then for any $N$ we would have $N\sum_{i=1}^{N}\frac{a_i}{b_i} \geq \frac{\sum_{i=1}^{N}a_i}{\sum_{i=1}^{N}b_i}$, but this is another story... $\endgroup$ – Mihail Poplavskyi Aug 31 '16 at 12:12
  • $\begingroup$ One more question: besides $a_1=a_2=\ldots=a_N$ or $b_1=b_2=\ldots=b_N$, what is the condition for the two terms to be equal? $\endgroup$ – wbchu Aug 31 '16 at 12:36
  • $\begingroup$ @wbchu, There are infinitely many examples even for $N=3$. Write an equation l.h.s.=r.h.s. and solve it in terms of $a_1$. You will see that besides of the case $\frac{2}{b_1}=\frac{1}{b_2}+\frac{1}{b_3}$ this equation has unique solution for any values of other parameters. $\endgroup$ – Mihail Poplavskyi Sep 1 '16 at 18:23
  • $\begingroup$ @Mihail Poplavskyi, I see. Thank you very much. $\endgroup$ – wbchu Sep 3 '16 at 3:13

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