I’ve learnt the AM–GM–HM inequality when I was a secondary school student.

\[ \frac1n \sum\limits_{i=1}^n x_i \ge \sqrt[n]{\prod_{i=1}^n x_i} \ge \frac{n}{\sum\limits_{i=1}^n x_i^{-1}} \]

But I didn’t know how to generalise it to integer power $s$. That is, if $k$ and $m$ are positive integers such that $k \le m$, then we have

\[ \left( \frac1n \sum\limits_{i=1}^n x_i^k \right)^{1/k} \le \left( \frac1n \sum\limits_{i=1}^n x_i^m \right)^{1/m}. \]

I read the proof in Wikipedia, and got stuck in the first step.

\[ \frac{\sum\limits_{i=1}^{n} w_{i}x_{i}^{k}}{\sum\limits_{i=1}^{n} w_{i}x_{i}^ {k-1}} \le \frac{\sum\limits_{i=1}^{n} w_{i}x_{i}^{k+1}}{\sum\limits_{i=1}^{n} w_{i}x_{i}^{k}} \]

I quickly realised that $w_i = 1/n$ in this case. Even though I know that the above inequality is eqivalent to

\[ \left( \sum\limits_{i=1}^n w_{i}x_{i}^{k} \right)^2 \le \left( \sum\limits_{i=1}^n w_{i}x_{i}^{k-1} \right) \left( \sum\limits_{i=1}^n w_{i}x_{i}^{k+1} \right). \]

At the first glance, I didn’t know how to relate this to the famous Cauchy–Schwartz inequality.

\[ \left( \sum\limits_{i=1}^n a_{i}b_{i} \right)^2 \le \left(\sum\limits_{i=1}^n a_{i}^2\right) \left(\sum\limits_{i=1}^n b_{i}^2\right) \]

Actually, setting $a_i = \sqrt{w_{i}x_{i}^{k-1}}$ and $b_i = \sqrt{w_{i}x_{i}^{k+1}}$ will do.