Why does the lower bound on the Hardy-Littlewood maximal function make it non-integrable?

Question

We have that $$Hf(x) \geq \frac{c}{|x|^n}$$ for some $c>0$ whenever $|x| \geq 1$. How does this lower bound show that that the maximal function is non-integrable? Perhaps if we could show that $\frac{c}{|x|^n}$ isn't integrable outside the unit ball we could show this. However, I am not sure how to do this. Similar questions seem to imply that it is obvious given the inequality.

Answer 1

Hint: Observe \begin{align} \int_{\mathbb{R}^n} \frac{dx}{|x|^n} =\int^\infty_0 \int_{|x|=r} \frac{1}{|x|^n}\ dS(x)dr \end{align}