Romanian Mathematical Olympiad

Denoting $y_1 = x_1$, $y_{k+1} = x_{k+1} - x_k$, for $k \in \{1, 2, \dots, n-1\}$, we get $x_k = y_1 + y_2 + \dots + y_k$, $\text{cu } y_1 > 0, y_k \ge 0, \forall k = \overline{2, n}$. With these notations, the inequality becomes $(a_1 + a_2 + \dots + a_n - c(b_1 + b_2 + \dots + b_n))y_1 + (a_2 + a_3 + \dots + a_n - c(b_2 + b_3 + \dots + b_n))y_2 + \dots + (a_n - cb_n)y_n \ge 0$ ($*$).

Plugging $y_1 = 1, y_2 = \dots = y_n = 0$, shows that necessarily $c \le \frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}$.

We show that, conversely, for every $c \le \frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}$ the inequality under consideration is true, so $\frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}$ is the required maximum.

We notice that $c \le \frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n} \le \frac{a_2+a_3+\dots+a_n}{b_2+b_3+\dots+b_n} \le \dots \le \frac{a_n}{b_n}$. Indeed, the inequality $\frac{a_k+a_{k+1}+\dots+a_n}{b_k+b_{k+1}+\dots+b_n} \le \frac{a_{k+1}+\dots+a_n}{b_{k+1}+\dots+b_n}$ can be written $a_k(b_{k+1} + \dots + b_n) \le b_k(a_{k+1} + \dots + a_n)$ and is justified by the inequalities $\frac{a_k}{b_k} \le \frac{a_{k+1}}{b_{k+1}}, \frac{a_k}{b_k} \le \frac{a_{k+2}}{b_{k+2}}, \dots, \frac{a_k}{b_k} \le \frac{a_n}{b_n}$.

This shows that every term of the left member of $(*)$ is at least 0, so $(*)$ is true.