How can I prove that,
$$\frac{a_{1}+a_{2}+\dots+a_{n}}{b_{1}+b_{2}+\dots+b_{n}} \le \max_i\left\{\frac{a_{i}}{b_{i}}\right\}$$
where $1 \le i \le n$, and $a_{i} \neq a_{j}$ and $b_{i} \neq b_{j}, \forall i \neq j$
Edit I have figured out that the above assumptions about $a_{i}$, and $b_{i}$ are not needed.
On
The $a_i$ can be arbitrary real numbers, but the $b_i$ need to be positive. Then $$ \tag{*} a_j \le b_j \cdot \left\{ \max_i \frac{a_i}{b_i} \right \} \quad \text{for $j =1, \ldots, n$} $$ and adding these gives the desired inequality.
If the $b_i$ are not required to be positive then the inequality must not hold, a counter-example is $$ \frac {2 - 1}{3 - 2} > \max \left\{ \frac{2}{3}, \frac{-1}{-2} \right \} \, . $$
Hint:
Suppose this holds for $n=2$ (prove this base case yourself). Then $$\frac{(a_1+...a_k)+a_{k+1}}{(b_1+...+b_k)+b_{k+1}}\le \max\left(\frac{a_1+...+a_k}{b_1+...+b_{k}}, \frac{a_{k+1}}{b_{k+1}}\right)$$