3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.