$$a_1b_1+a_2b_2 + \cdots +a_nb_n = (a_1 -a_2)\times \frac{b_1}{1}+2(a_2-a_3) \times \frac{b_1+b_2}{2} + \cdots + na_n \times \frac{b_1+b_2+ \cdots +b_n}{n} \le (a_1+a_2+ \cdots a_n)\text{max}\{ \frac{b_1}{1}, \frac{b_1+b_2}{2}, \cdots, \frac{b_1+b_2+ \cdots +b_n}{n} \}$$using abel summation will make this problem done...

Actually, positivity of $b$s are not necessary. We will prove the following statement. For every positive integer $n$, for every positive reals $a_1 \ge a_2 \ge \ldots \ge a_n$, for every reals $b_1$, $b_2$, $\ldots$, $b_n$, the following holds. $$\frac{a_1b_1+a_2b_2 + \cdots +a_nb_n}{a_1+a_2+ \cdots a_n} \le \text{max}\{ \frac{b_1}{1}, \frac{b_1+b_2}{2}, \cdots, \frac{b_1+b_2+ \cdots +b_n}{n} \}$$ To prove this, suppose that for some integer $n$, there are positive reals $a_1 \ge a_2 \ge \ldots \ge a_n$, so that there are $b_1$, $b_2$, $\ldots$, $b_n$, the following holds; $$\frac{a_1b_1+a_2b_2 + \cdots +a_nb_n}{a_1+a_2+ \cdots a_n} > \text{max}\{ \frac{b_1}{1}, \frac{b_1+b_2}{2}, \cdots, \frac{b_1+b_2+ \cdots +b_n}{n} \}$$Since the inequality is invariant under parallel transforation for $b$s, i.e. $b_j \mapsto b_j+\alpha$, we may assume that $\sum_{j=1}^n a_j b_j =0$. Then we have $B_j:= b_1 + b_2 + \cdots + b_j < 0$ for each $j = 1,2, \ldots , n$. Applying Abel summation formula, we have $$ 0 = \sum a_j b_j = B_n a_n - \sum_{k=1}^{n-1} B_k (a_{k+1}-a_k) <0 $$which is an absurdity. Therefore, we should have the statement.