China Hong Kong Mathematical Olympiad

By the Cauchy-Schwarz inequality, we have
$$(1^2 a_1 + 2^2 a_2 + \dots + k^2 a_k)(a_1^{-1} + a_2^{-1} + \dots + a_k^{-1}) \ge (1 + 2 + \dots + k)^2 = \frac{k^2 (k+1)^2}{4}.$$
This implies
$$\frac{k}{a_1^{-1} + a_2^{-1} + \dots + a_k^{-1}} \le \frac{4}{k(k+1)^2} \sum_{j=1}^{k} j^2 a_j.$$

Also, using $\frac{1}{k^2} - \frac{1}{(k+1)^2} = \frac{2k+1}{k^2(k+1)^2} \ge \frac{2}{k(k+1)^2}$, we obtain
$$\begin{align*} \sum_{k=1}^{n} \frac{k}{a_1^{-1} + a_2^{-1} + \cdots + a_k^{-1}} &\le \sum_{k=1}^{n} \sum_{j=1}^{k} j^2 a_j \frac{4}{k(k+1)^2} \\ &\le 2 \sum_{k=1}^{n} \sum_{j=1}^{k} j^2 a_j \left( \frac{1}{k^2} - \frac{1}{(k+1)^2} \right) \\ &= 2 \sum_{j=1}^{n} \sum_{k=j}^{n} j^2 a_j \left( \frac{1}{k^2} - \frac{1}{(k+1)^2} \right) \\ &= 2 \sum_{j=1}^{n} j^2 a_j \left( \frac{1}{j^2} - \frac{1}{(n+1)^2} \right) \\ &< 2 \sum_{j=1}^{n} a_j. \end{align*}$$
This proves the desired inequality.

Consider $a_j = \frac{1}{j}$ for each $j$. Then
$$\sum_{k=1}^{n} \frac{k}{a_1^{-1} + a_2^{-1} + \cdots + a_k^{-1}} = \sum_{k=1}^{n} \frac{k}{1 + 2 + \cdots + k} = \sum_{k=1}^{n} \frac{2}{k + 1}$$
and $\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} \frac{1}{k}$. For any $c$ such that $0 < c < 2$, we have
$$c \sum_{k=1}^{n} a_k - \sum_{k=1}^{n} \frac{k}{a_1^{-1} + a_2^{-1} + \cdots + a_k^{-1}} = c + (c-2) \sum_{k=2}^{n} \frac{1}{k} - \frac{2}{n+1} < 0$$
for sufficiently large $n$, since the harmonic series $\sum_{k=2}^{\infty} \frac{1}{k}$ diverges to positive infinity. This shows we cannot replace the number 2 by any smaller constant $c$.