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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?

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    $\frac{1}{n}\sum_{i=1}^n \frac{y_i}{x_i} \geq 1$ by AM-GM – 2017-02-20
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    @Henry sorry, fixed now. How do we use AM-GM to get that? – 2017-02-20
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    If LHS is AM, what is GM? Also can be done easily by rearrangement. – 2017-02-20
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    Follows immediately from [this lemma](http://math.stackexchange.com/questions/691807/proofs-of-am-gm-inequality/1665056#1665056). – 2017-02-20

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

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    *sigh* I'm feeling particularly dumb right now. Thank you, I can't believe it was that obvious... – 2017-02-20
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    @ZainPatel don't think yourself dumb — the beauty in great mathematics is that once you see how it works it seems so simple :) – 2017-02-20
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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 $(1, \ldots, n)$ $$ \sum_{i=1}^n \frac{y_i}{x_i} = \sum_{i=1}^n x_{\sigma(i)} \frac{1}{x_i} \ge \sum_{i=1}^n x_i \frac{1}{x_i} = n \, . $$