Suppose we have a positive sequence $\{x_i\}$ where $i=1, 2, \dots , n$. And let $m=\mathrm{min}(x_i), \; M = \mathrm{max}(x_i)$. Notice that $m, M>0$.
Prove this inequality (discrete version):
$$ \frac{\sum_{i=1}^{n} x_i }{n} \cdot \frac{ \sum_{i=1}^{n} \frac{1}{x_i} }{n} \le \frac{m+M}{2} \cdot \frac{ \frac{1}{m} + \frac{1}{M} }{2} \;.$$
Or you can prove this continuous version, where $f(x)$ is positive and continuous on $[0, 1]$ , and $m$ and $M$ are min and max values of $f(x)$ on $[0, 1]$, respectively:
$$ \int_0^1 f(x) \,\mathrm{d}x \int_0^1 \frac{1}{f(x)} \,\mathrm{d}x \le \frac{m+M}{2} \cdot \frac{ \frac{1}{m} + \frac{1}{M} }{2} \;.$$
Important: I do have a proof, because the continuous version is actually an excercise of a workbook of mine. But that proof is too tricky, lacking insight as to why the product of two types of "average"s is controlled by their max and min values. So I brought it up here in seeking of an insightful solution. Hopefully finding a proof directly to the discrete version, for it is easier to imagine with my brain.
Note: This question already has answers here Reverse Cauchy Schwarz for integrals. But that is more of a general case. I am looking for a more insightful and easy-to-understand proof with regard to this specific situation where the $f(x), g(x)$ in the original answer is taken as $f(x), \frac{1}{f(x)}$.
