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 
 A: 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
$$\frac{1}{1} + \ldots + \frac{1}{n} = \sum_{k=1}^n \frac{1}{k} \leq \sum_{k=1}^n \frac{a_k}{k^2} = \frac{a_1}{1^2} + \ldots + \frac{a_n}{n^2}$$
A: First we quote (part of) the  Rearrangement Inequality, in the notation used by  Wikipedia: 
$$x_ny_1 + \cdots + x_1y_n \le x_{\sigma (1)}y_1 + \cdots + x_{\sigma (n)}y_n\tag{1} $$
for every choice of real numbers
$$ x_1\le\cdots\le x_n\quad\text{and}\quad y_1\le\cdots\le y_n,$$
and every permutation
$$ x_{\sigma(1)},\dots,x_{\sigma(n)}\,$$ 
We can assume that the $a_i$ are a permutation of $\{1,2,\dots,n\}$.  For if that is not the case, and  we replace the smallest of the $a_i$ by $1$, the second smallest by $2$, and so on, we weaken the inequality. 
We want to prove that
$$\frac{a_n}{n^2}+\frac{a_{n-1}}{(n-1)^2}+\cdots +\frac{a_1}{1^2}\ge \frac{1}{n}+\frac{1}{n-1}+\cdot +\frac{1}{1}.\tag{2}$$
Let $y_i=\frac{1}{(n+1-i)^2}$ and let $x_i=i$. Then the Rearrangement Inequality (1) directly gives (2).
