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](https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_induction). 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.