Eighteenth STARS OF MATHEMATICS Competition

Let $A_j = \{i: 2^{j-1} \le a_i < 2^j\}, j = 1, 2, \dots, 11$. The $A_j$ form a partition of the index set $\{1, 2, \dots, n\}$.

Note that, for every $j$ in the range 1 through 11, if $A_j$ is non-empty, then
$$\begin{aligned} \sum_{i \in A_j} \frac{1}{a_i} (a_1 + a_2 + \dots + a_i) &= \sum_{i \in A_j} \left( \frac{a_1 + a_2 + \dots + a_{i-1}}{a_i} + 1 \right) \\ &> \sum_{k=0}^{|A_j|-1} \frac{k \cdot 2^{j-1}}{2^j} + |A_j| = \frac{1}{2} \sum_{k=0}^{|A_j|-1} k + |A_j| \\ &= \frac{1}{4} |A_j|(|A_j| - 1) + |A_j| = \frac{1}{4} |A_j|(|A_j| + 3); \end{aligned}$$

As the $A_j$ form a partition of the index set $\{1, 2, \dots, n\}$ and at least one $A_j$ is non-empty,
$$$$
$$\begin{aligned} \sum_{i=1}^{n} \frac{1}{a_i} (a_1 + a_2 + \dots + a_i) &= \sum_{j=1}^{11} \sum_{i \in A_j} \frac{1}{a_i} (a_1 + a_2 + \dots + a_i) \\ &> \frac{1}{4} \sum_{j=1}^{11} |A_j|(|A_j| + 3) = \frac{1}{4} \left( \sum_{j=1}^{11} |A_j|^2 + 3n \right) \\ &\ge \frac{1}{4} \left( \frac{n^2}{11} + 3n \right) = \frac{1}{44} n(n + 33), \text{ by Cauchy-Schwarz.} \end{aligned}$$