China Mathematical Competition (Complementary Test)

Given $j$ ($1 < j < n$), the number of ternary groups $(i, j, k)$ satisfying $1 \le i < j < k \le n$ and
$$\frac{a_j - a_i}{a_k - a_j} = r \qquad \textcircled{1}$$
is denoted as $g_j(r)$. For fixed $i, j$ with $i < j$, there is at most one $k$ satisfying ①; so there are $j-1$ ways to choose $i$, which means $g_j(r) \le j-1$. In a similar way, for fixed $j, k$ with $k > j$, there is at most one $i$ satisfying ①; so there are $n-j$ ways to choose $k$, which means $g_j(r) \le n-j$. Therefore,
$$g_j(r) \le \min\{j-1, n-j\}.$$
Then, when $n$ is even (i.e., $n = 2m$), we have
$$\begin{align*} f_n(r) &= \sum_{j=2}^{n-1} g_j(r) = \sum_{j=2}^{m-1} g_j(r) + \sum_{j=m}^{2m-1} g_j(r) \\ &\le \sum_{j=2}^{m} (j-1) + \sum_{j=m+1}^{2m-1} (2m-j) = \frac{m(m-1)}{2} + \frac{m(m-1)}{2} \\ &= m^2 - m < m^2 = \frac{n^2}{4}. \end{align*}$$
When $n$ is odd (i.e. $n = 2m + 1$), we have
$$\begin{aligned} f_n(r) &= \sum_{j=2}^{n-1} g_j(r) = \sum_{j=2}^{m} g_j(r) + \sum_{j=m+1}^{2m} g_j(r) \\ &\le \sum_{j=2}^{m} (j-1) + \sum_{j=m+1}^{2m} (2m+1-j) \\ &= m^2 < \frac{n^2}{4}. \end{aligned}$$
The proof is completed. $\Box$