shorter proof of generalized mediant inequality?

Question

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?

Answer 1

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.