$\sum_{i=1}^n \frac{y_i}{x_i} \geq n$ where $y_i$ is a reordering of $x_i$

Question

The first part of the question had us prove that if $x_1,\ldots, x_n$ are positive real numbers then $\displaystyle \left(\frac{1}{n}\sum_{i=1}^n \frac{1}{x_i}\right)^{-1} \leq \frac{1}{n}\sum_{i=1}^n x_i$. I've done this by using AM-GM on $\frac{1}{x_i}$.

Now, I want to show that, if $y_1, \ldots, y_n$ is any reordering of $x_1,\ldots,x_n$ then $$\frac{1}{n}\sum_{i=1}^n \frac{y_i}{x_i} \geq 1$$

The fact that there's $\frac{1}{x_i}$ in there makes me think of a link to the first part, but I can't find anyway to use. I thought about trying something like $\sum \frac{y_i}{x_i} \geq \frac{1}{\max(x_i)} \sum x_i \geq n^2 (\sum x_i^{-1})^{-1}$ but there's no way to get a bound on the last term. Any help?

Answer 1

$$\displaystyle \frac{1}{n}\sum_{i=1}^n \frac{y_i}{x_i} \geq \sqrt[n]{\prod_{i=1}^n \frac{y_i}{x_i} } =\sqrt[n]{ \frac{\prod_{i=1}^n y_i}{\prod_{i=1}^n x_i} } =\sqrt[n]{1} =1$$ by the AM-GM inequality if all the values are positive

Answer 2

The conclusion also follows from the Rearrangement inequality. Without loss of generality we can assume that $$ x_1 \le x_2 \le \ldots \le x_n $$ and therefore $$ \frac{1}{x_1} \ge \frac{1}{x_2} \ge \ldots \ge \frac{1}{x_n} $$ Then for some permutation $\sigma$ of $(1, \ldots, n)$ $$ \sum_{i=1}^n \frac{y_i}{x_i} = \sum_{i=1}^n x_{\sigma(i)} \frac{1}{x_i} \ge \sum_{i=1}^n x_i \frac{1}{x_i} = n \, . $$