Taiwan IMO Selection Camp

The original problem asks to prove that: for every $m \in \{1, 2, \dots, n\}$,
$$\frac{1}{m} \sum_{k=1}^{m} \frac{a_k}{b_k} \geq \frac{M}{N}, \text{ where } n = 107.$$

Let $\sum_{k=1}^{n} a_k = a \ge M$ and $\sum_{k=1}^{n} b_k = b \le N$, and define
$$x_k = \frac{a_k}{a}, \quad y_k = \frac{b_k}{b}, \quad k = 1, 2, \dots, n.$$

Then we have $\sum_{k=1}^{n} x_k = \sum_{k=1}^{n} y_k = 1$, and the sequence $\{x_k\}$ is decreasing, while the sequence $\{y_k\}$ is increasing. Therefore, the sequence $\{\frac{x_k}{y_k}\}$ is decreasing, and
$$\frac{x_1}{y_1} \ge 1 \ge \frac{x_n}{y_n}.$$
We may take $k_0 \in \{1, 2, \dots, n\}$ such that
$$\frac{x_1}{y_1} \ge \frac{x_2}{y_2} \ge \cdots \ge \frac{x_{k_0}}{y_{k_0}} \ge 1 \ge \frac{x_{k_0+1}}{y_{k_0+1}} \ge \cdots \ge \frac{x_n}{y_n}.$$

(1) For a positive integer $m \in \{1, 2, \dots, k_0\}$, we have
$$\sum_{k=1}^{m} \frac{x_k}{y_k} \geq \sum_{k=1}^{m} \frac{y_k}{x_k} = m;$$

$$\sum_{k=1}^{m} \frac{a_k}{b_k} = \sum_{k=1}^{m} \frac{ax_k}{by_k} \geq \frac{ma}{b} \geq \frac{mM}{N},$$

$$\frac{1}{m} \sum_{k=1}^{m} \frac{a_k}{b_k} \geq \frac{M}{N}.$$

(2) For a positive integer $m \in \{k_0 + 1, k_0 + 2, \dots, n\}$, from
$$\sum_{k=1}^{n} x_k = \sum_{k=1}^{n} y_k,$$
we have
$$\sum_{k=1}^{k_0} (x_k - y_k) = \sum_{k=k_0+1}^{n} (y_k - x_k) \geq \sum_{k=k_0+1}^{m} (y_k - x_k).$$

$$\begin{aligned} \sum_{k=1}^{m} \frac{x_k}{y_k} &= \sum_{k=1}^{k_0} \left(1 + \frac{x_k - y_k}{y_k}\right) + \sum_{k=k_0+1}^{m} \left(1 - \frac{y_k - x_k}{y_k}\right) \\ &\geq m + \frac{1}{y_{k_0+1}} \left(\sum_{k=1}^{k_0} (x_k - y_k) - \sum_{k=k_0+1}^{m} (y_k - x_k)\right) \geq m. \end{aligned}$$

$$\sum_{k=1}^{m} \frac{a_k}{b_k} = \sum_{k=1}^{m} \frac{ax_k}{by_k} \geq \frac{ma}{b} \geq \frac{mM}{N},$$

$$\frac{1}{m} \sum_{k=1}^{m} \frac{a_k}{b_k} \geq \frac{M}{N}.$$