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.

Hint: Suppose $0<a<b$ and $0<x<y.$ Then $a/x + b/y < b/x + a/y.$
