Let $p, q \in \mathbb{Z}$. Assuming the lemma that when $p > 0$ then $M_p > G$, I need to prove that if $q < 0$ then $M_q < G$.
I'm trying to prove it this way:
Let $q = -k$ and $k > 0$. Then $(M_k)^k > G^k$ then $(M_{-q})^{-q} > G^{-q}$ then $G^q > (M_{-q})^q$ then $G > M_{-q}$, but now that negative subindex is bothering me and I can't find a way to start with a hypothesis that allows me to conclude that $G > M_{q}$.
I found a solution here, in the end of the article, but I don't know how is this implication correct:
$G^{-q} < (M_{-q})^{-q} \implies G^q > M_{q}^q$
Shouldn't $G^{-q} < (M_{-q})^{-q}$ just imply $G^q > M_{-q}^q$? How can I prove this?
PD: For context, this is the problem 20 of Apostol's Calculus, volume I first chapter, so I would like to keep the proof as basic as possible. Thanks!
Let $p=-q>0$, and apply your lemma to $\frac{1}{x_i}$: $$\left(\frac{\sum\left(\frac{1}{x_i}\right)^{p}}{n}\right)^{\frac 1p} \geqslant \left(\Pi \frac{1}{x_i}\right)^{1/n} \implies \left(\frac{\sum\left(\frac{1}{x_i}\right)^{-q}}{n}\right)^{\frac {-1}{q}} \geqslant \left(\Pi \frac{1}{x_i}\right)^{1/n}\\ \implies \left(\Pi x_i\right)^{1/n} \geqslant \left(\frac{\sum x_i^q}{n}\right)^{\frac {1}{q}} $$