Prove that $\sum_{i = 1}^n \frac{x_i}{i^2} \geq \sum_{i = 1}^n \frac{1}{i}.$ 
Let $x_1,x_2,\ldots,x_n$ be distinct positive integers. Prove that $$\displaystyle \sum_{i = 1}^n \dfrac{x_i}{i^2} \geq \sum_{i = 1}^n \dfrac{1}{i}.$$

Attempt
I tried using Cauchy-Schwarz and I got that $$(x_1^2+x_2^2+\cdots+x_n^2) \left (\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+ \dfrac{1}{n^2} \right ) \geq \left ( \dfrac{x_1}{1}+\dfrac{x_2}{2}+\cdots+\dfrac{x_n}{n} \right)^2 = \displaystyle \left (\sum_{i = 1}^n \dfrac{x_i}{i} \right)^2,$$ but this doesn't seem to help.

Let $0 < y_1 < y_2 < \ldots < y_n$ be an rearrangement of $x_1, x_2, \ldots x_n$ sorted in ascending order.  
Since $y_k$ are distinct positive integers, one can show $y_k \ge k$ for $k = 1,\ldots, n$ by induction.
Since the finite sequence $\frac{1}{1^2}, \frac{1}{2^2}, \ldots, \frac{1}{n^2}$ is ascending, by rearrangement inequality, we have
$$\sum_{k=1}^n \frac{x_k}{k^2} \ge \sum_{k=1}^n \frac{y_k}{k^2} \ge \sum_{k=1}^n \frac{k}{k^2} = \sum_{k=1}^n \frac{1}{k}$$
