real analysis - Lim Sup and Lim Inf relations between the root test and the ratio test

I'm trying to solve the following problem and while looking around I found a couple of notes (Link) by Pete L. Clark where it's discussed but can't really comprehend his proof.

The problem:
Prove that $ \lim\inf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} \leq \lim\inf\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$

His proof, as I understood it, is as follows:
Let $r > \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} $, so there's an $n_0$ so that for all $k \geq 1$ then $ \frac{A_{n_0+k}}{A_{n_0 + k - 1}} < r $.

From that we get $ A_{n_0 + k} < r A_{n_0+k-1} $ and that implies $A_{n_0 + k} < r^{k} A_{n_0} $. Rewriting that as $A_{n_0+k} ^{\frac{1}{n_0+k}} < r (\frac{A_{n_0}}{r^n})^{\frac{1}{n_0+k}}$ and by letting $ k \rightarrow \infty$ we see that $\lim\sup\limits_{k\rightarrow \infty} A_{n_0+k} ^{\frac{1}{n_0+k}}$ is at most $r$.

So, we have that $r > \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} $ and $ r \geq \lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n}$

He says that that implies $\lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq
\lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$, but I really don't see how that's the case to be honest. And I'm also not too sure how the analogous argument for $\lim\inf$ would look like.

I'm sure the proof is correct, as I've seen a lot of people recommend that particular set of notes, but I just can't seem to grasp it and it looks so simple.

Any help, both with this particular proof and with any other that applies, would be greatly appreciated.

Answer

Try to prove this:

Suppose that for each $r>b$, we have $a\leqslant r$. Then $a\leqslant b$.

Equivalently

Suppose that for each $\epsilon >0$, we have $y\leqslant \epsilon$. Then $y \leqslant 0$.

Pete (who is a user here!) is using this with $b=\limsup\limits_{n\to\infty} \dfrac{ A_{n+1}}{A_n}$ and $a=\limsup\limits_{n\to\infty} A_n^{1/n}$.

For the argument using $\liminf$; you start with $\liminf\limits_{n\to\infty}\frac{A_{n+1}}{A_n}$ and choose $r$ smaller than this. The steps are the same but with signs reversed. Then you use

Suppose that for each $r