Deduce inequality between geometric mean and power mean from AM-GM

Question

Below is exercise 20 from Tom Apostol's "Calculus" Vol. 1 (2nd edition). I need help solving part (b).

The geometric mean $G$ of $n$ positive real numbers $x_1, x_2, ..., x_n$ is defined by the formula $G = (x_1x_2,...x_n)^{1/n}$.

(a) Let $M_p$ denote the pth power mean. Prove that $G \le M_1$ and that $G = M_1$ only when $x_1 = x_2 = ... = x_n$.

(b) Let $p$ and $q$ be integers, $q \lt 0 \lt p$. From part (a) deduce that $M_q \lt G \lt M_p$ when $x_1, x_2, ..., x_n$ are not all equal.

One of the previous exercises introduced the below theorem, which looks like something that may be used here, but I'm not sure exactly how. I know that every $q \gt 1$ satisfies $q \gt 2p$ for some $p$, but that would still require to show that either $M_q \gt M_{2p}$ or $M_q \gt G$.

If $p \gt 0$, $M_p \lt M_{2p}$ when $x_1, x_2,..., x_n$ are not all equal.

This exercise is hard! I've spent 3 hours doing part (a) and didn't manage to do it alone, but found solution in Wikipedia article. Still have no idea how to do part (b).

Assuming that this comes from Introduction to Calculus book, please don't use any advanced theorems. Ideally, if you have the book, consult it to know what theorems were introduced so far and are available to use.

Answer 1

Hint: write the AM-GM inequality for numbers $x_1^p, x_2^p, \ldots, x_n^p$. Raise the inequality to the power $\frac 1p$.

Answer 2

The point b) is a consequence of the fact that power mean function $M_p$ is strictly increasing if $x_1, x_2, \dots$ are not equal.

If you take the derivative of $M_p$, i.e., $\frac{d}{dp}\left(\sum_iw_ix_i^p\right)^{1/p}$, it is not complicated to show that the derivative is stricly positive (using e.g. Cauchy-Schwarz inequality) and thus $M_p$ is increasing function of $p$.

That knowing, it follows that for any $q<p$, $M_q<M_p$.

Also the geometric mean is just a special case when $\lim_{p\to0}M_p=M_0=G$.

And thus also if $ q<0<p$, then $M_q<M_0<M_p$.