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If $\frac{a_1}{b_1}$, $\frac{a_2}{b_2}$, $\frac{a_3}{b_3}$, $\dots, \frac{a_n}{b_n}$ are unequal fractions. All numbers are positive.

Then the ratio:

$$\frac{(a_1+a_2+\dots+a_n)}{(b_1+b_2+\dots+b_n)}$$

will lie between the lowest and highest of these fractions.

How do I prove this ?

Any hints??

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    $\begingroup$ have you tried induction? $\endgroup$ – J. W. Tanner May 10 at 15:45
  • $\begingroup$ @J.W.Tanner I have no clue on how to do induction. Could you help ? $\endgroup$ – ng.newbie May 10 at 15:45
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    $\begingroup$ In the question you should mention $b_i$ to be positive for all i, otherwise consider $ 1/1$ And $-4/-2=2$. $\endgroup$ – Nabakumar Bhattacharya May 10 at 15:54
  • $\begingroup$ @NabakumarBhattacharya Yes I will do that. $\endgroup$ – ng.newbie May 10 at 15:55
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Suppose the highest one is $\frac{a_n}{b_n}$ and the lowest one is $a_1/b_1$ And the denominators are positive. See, $b_n(a_1+...+a_n)\leq a_n(b_1+..+b_n)$ since $a_i/b_i \leq a_n/b_n \Rightarrow b_na_i \leq a_n b_i \ \forall i$. Similarly you can do the other side.

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  • $\begingroup$ I did not understand this step : $b_n(a_1+...+a_n)\leq a_n(b_1+..+b_n)$. How can you say that ? $b_na_i \leq a_n b_i$, is for a single $i$ not a summation. $\endgroup$ – ng.newbie May 10 at 15:57
  • $\begingroup$ That’s true for all i now take summation. $\endgroup$ – Nabakumar Bhattacharya May 10 at 15:58
  • $\begingroup$ Did not understand - how is $(a_1+...+a_n)$ equivalent to $a_i$ ? $\endgroup$ – ng.newbie May 10 at 16:00
  • $\begingroup$ You have the last inequality for all i, now what will happen if you add them all? $\endgroup$ – Nabakumar Bhattacharya May 10 at 16:01
  • $\begingroup$ Yes it is for all i, INDIVIDUALLY. The meaning changes when I add them together. Am I correct ? $\endgroup$ – ng.newbie May 10 at 16:03

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