What is the most elementary proof of these inequalities? Let $p$ be a non-zero integer, and let $x_1$, $\ldots$, $x_n$ be $n$ positive real numbers. Then we define the $p$-th power mean $M_p$ of these numbers as $$ M_p \colon= (\frac{x_1^p + \ldots + x_n^p}{n})^{1/p}. $$ 
Also, the geometric mean of these numbers is defined as $$ G \colon= \sqrt[n]{x_1\ldots x_n}.$$ 
Let $p$ and $q$ be integers such that $q < 0 < p$. Then how to establish the following inequalities using only the most elementary ideas? $$M_q < G < M_p$$, where $x_1$, $\ldots$, $x_n$ are not all equal. 
 A: We can show that $M_q\le G$ and $G\le M_q$ are just equivalent, and both are equivalent to the AM-GM inequality.
1) If $q<0$ then let $p=-q >0$ and the inequality $M_q \le G$ becomes
$$\frac{1}{\left(\frac{\frac{1}{x_1^p}+\cdots+\frac{1}{x_n^p}}{n}\right)^{1/p}} \le \sqrt[n]{x_1\ldots x_n},$$
which is equivalent to 
$$\frac{1}{\sqrt[n]{x_1\ldots x_n}} \le \left(\frac{\frac{1}{x_1^p}+\cdots+\frac{1}{x_n^p}}{n}\right)^{1/p}.$$
Denote $y_i=\frac{1}{x_i}$, then the last inequality becomes
$$\sqrt[n]{y_1\ldots y_n} \le \left(\frac{y_1^p+\cdots+y_n^p}{n}\right)^{1/p},$$
which is $G \le M_p$ with variables $y_1,\ldots,y_n$.
2) Now we show that $G \le M_p$ is equivalent to the AM-GM inequality.
$$\sqrt[n]{x_1\ldots x_n} \le \left(\frac{x_1^p+\cdots+x_n^p}{n}\right)^{1/p} \Leftrightarrow \sqrt[n]{(x_1\ldots x_n)^p} \le \frac{x_1^p+\cdots+x_n^p}{n}.$$
Denote $y_i=x_i^p$ then the last inequality becomes 
$$\sqrt[n]{y_1\ldots y_n} \le \frac{y_1+\cdots+y_n}{n},$$
which is the AM-GM inequality.
3) For an elementary proof for the AM-GM inequality, I recommend the first proof by induction in this article. Nice and simple idea.  
