# How do I prove this property for ratio?

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??

• have you tried induction? – J. W. Tanner May 10 at 15:45
• @J.W.Tanner I have no clue on how to do induction. Could you help ? – ng.newbie May 10 at 15:45
• In the question you should mention $b_i$ to be positive for all i, otherwise consider $1/1$ And $-4/-2=2$. – Nabakumar Bhattacharya May 10 at 15:54
• @NabakumarBhattacharya Yes I will do that. – ng.newbie May 10 at 15:55

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.
• 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. – ng.newbie May 10 at 15:57
• Did not understand - how is $(a_1+...+a_n)$ equivalent to $a_i$ ? – ng.newbie May 10 at 16:00