Rearrangement and Cauchy Let $a_1, \ldots, a_n$ be distinct positive integers. I want to prove that
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + \cdots + \frac{a_n}{n^2} \geq \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}.$$
I've been considering using Rearrangement and Cauchy Schwarz, but cannot make any progress 

Use the following: If $a_1, \ldots, a_n$ are distinct positive integers, then we can sort them according to their size $a_{n_1} \leq a_{n_2} \leq \ldots \leq a_{n_n}$ and get
$$k \leq a_{n_k} ~ \forall k = 1, \ldots, n.$$
Hence, using Cauchy Schwarz, we compute
\begin{align*}
\left( \sum_{k=1}^n \frac{1}{k} \right)^2 & = \left( \sum_{k=1}^n \frac{\sqrt{a_k}}{k}  \cdot \frac{1}{\sqrt{a_k}}\right)^2 \leq \left( \sum_{k=1}^n \frac{a_k}{k^2} \right) \cdot \left( \sum_{k=1}^n \frac{1}{a_k} \right) = \left( \sum_{k=1}^n \frac{a_k}{k^2} \right) \cdot \left( \sum_{k=1}^n \frac{1}{a_{n_k}} \right)\\
& \leq \left( \sum_{k=1}^n \frac{a_k}{k^2} \right) \cdot \left( \sum_{k=1}^n \frac{1}{k} \right)
\end{align*}
Thus, dividing the inequality by $\sum_{k=1}^n 1/k$ yields
$$\frac{1}{1} + \ldots + \frac{1}{n} = \sum_{k=1}^n \frac{1}{k} \leq \sum_{k=1}^n \frac{a_k}{k^2} = \frac{a_1}{1^2} + \ldots + \frac{a_n}{n^2}$$
