Here's Theorem 3.37 in the book Principles of Mathematical Analysis by Walter Rudin, third edition:
For any sequence $\{c_n\}$ of positive numbers, $$\lim_{n\to\infty} \inf \frac{c_{n+1}}{c_n} \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ $$ \lim_{n\to\infty} \sup \sqrt[n]{c_n} \leq \lim_{n\to\infty} \sup \frac{c_{n+1}}{c_n}.$$
Now Rudin has given a proof of the second inequality. Here's my proof of the first.
Let $$\alpha = \lim_{n\to\infty} \inf \frac{c_{n+1}}{c_n}.$$ Then $\alpha \geq 0$. If $\alpha = 0$, then we're done since $$\lim_{n\to\infty} \sqrt[n]{c_n} \geq 0.$$ So we suppose that $\alpha > 0$ and choose a real number $\beta$ such that $0 < \beta < \alpha$. Then by the result analogous to Theorem 3.17 (b), there is an integer $N$ such that $n \geq N$ implies $$ \frac{c_{n+1}}{c_n} > \beta,$$ which in turn implies $$c_{n+1} > \beta c_n.$$ So for each $n \geq N$, we have $$c_n \geq \left( c_N \cdot \beta^{-N} \right) \cdot \beta^n. $$ Thus, for $n \geq N$, we have $$\sqrt[n]{c_n} \geq \beta \sqrt[n]{ c_N \cdot \beta^{-N} }.$$
Then taking the limit inferior of both sides, we get $$\lim_{n\to\infty}\inf \sqrt[n]{c_n} \geq \lim_{n\to\infty} \inf \left( \beta \sqrt[n]{ c_N \beta^N} \right) = \lim_{n\to\infty} \left( \beta \sqrt[n]{ c_N \beta^N} \right) = \beta \cdot 1 = \beta$$ by Theorem 3.20 (b).
Thus we have shown that for any (positive) real number $\beta$ such that $\beta < \alpha$, we have $$\beta \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ which implies that $$\alpha \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ as required.
Is the above proof correct? If so, then is my presentation good enough? If not, then where does the problem lie?