shorter proof of generalized mediant inequality?

Show $\frac{a_{1}+...+a_{n}}{b_{1}+...+b_{n}}$ is between the smallest and largest fraction $\frac{a_{i}}{b_{i}}$, where $b_{i}>0$.

Attempt

Assume the largest is $\frac{a_{n}}{b_{n}}\Rightarrow$

$\frac{a_{n}}{b_{n}}-\frac{a_{1}+...+a_{n}}{b_{1}+...+b_{n}}\Rightarrow $

$\frac{b_{1}+...+b_{n-1}}{{b_{1}+...+b_{n}}}[\frac{a_{n}}{b_{n}}-\frac{a_{1}}{b_{1}+...+b_{n-1}}-...-\frac{a_{n-1}}{b_{1}+...+b_{n-1}}]\Rightarrow $

if $a_{1}<0$, $w=\frac{a_{n}}{b_{n}}-\frac{a_{1}}{b_{1}+...+b_{n-1}}-...-\frac{a_{n-1}}{b_{1}+...+b_{n-1}}>\frac{a_{n}}{b_{n}}-\frac{a_{2}}{b_{1}+...+b_{n-1}}-...-\frac{a_{n-1}}{b_{1}+...+b_{n-1}}$

any hints or solutions?

Solution 1:

If you are willing to grant that the weighted average of several terms falls between the smallest term and the largest term (pretty easily shown), we can do the proof in one line.

$\frac{a_1+...+a_n}{b_1+...+b_n}$ = ($\frac{a_1}{b_1}$)($\frac{b_1}{b_1+...b_n}$) +...+($\frac{a_n}{b_n}$)($\frac{b_n}{b_1+...b_n}$) This is a weighted average of the $\frac{a_i}{b_i}$ terms and so must not be greater than the largest or smaller than the smallest.

Solution 2:

We have that

$$\frac{a_n}{b_n}\ge \frac{a_i}{b_i}\iff a_nb_i\ge a_ib_n\;\;,\;\;\forall i\implies$$

$$\frac{a_n}{b_n}\ge\frac{a_1+\ldots+a_n}{b_1+\ldots+b_n}\iff a_nb_1+\ldots +a_nb_n\ge a_1b_n+\ldots+a_nb_n$$

and the claim follows from the fist part above

Now you try the other inequality.