Japan Mathematical Olympiad

Let $X = \sum_{i=1}^{n} i a_{i}$, $Y = \sum_{i=1}^{n} \frac{a_{i}}{i}$. We have to find the maximum possible value of the quantity $X Y^{2}$.

First, we note that for each $i \in \{1, 2, \dots, n\}$, $i + \frac{n}{i} \le n + 1$ holds. This follows since $(n+1) - (i + \frac{n}{i}) = \frac{1}{i}(i-1)(n-i) \ge 0$. Using this fact we get
$$X + nY = \sum_{i=1}^{n} \left(i + \frac{n}{i}\right) a_i \le \sum_{i=1}^{n} (n+1)a_i = n+1.$$
We then apply the inequality on additive and multiplicative means to the three quantities $X$, $\frac{nY}{2}$, $\frac{nY}{2}$ to obtain
$$X \cdot \frac{nY}{2} \cdot \frac{nY}{2} \le \left(\frac{X+nY}{3}\right)^3 \le \left(\frac{n+1}{3}\right)^3 = \frac{(n+1)^3}{27},$$
from which we conclude that $XY^2 \le \frac{4(n+1)^3}{27n^2}$ holds.

On the other hand, by choosing
$$a_1 = \frac{2n-1}{3(n-1)}, \quad a_2 = \dots = a_{n-1} = 0, \quad a_n = \frac{n-2}{3(n-1)}$$
we get $XY^2 = \frac{4(n+1)^3}{27n^2}$, which implies that $\frac{4(n+1)^3}{27n^2}$ is the desired maximum value.