Inequality between limits inferior Let $(a_n)$ be a non-zero real sequence with $\left (\frac{a_{n+1}}{a_n} \right)$ bounded. How might we prove that
$$\liminf_{n\to\infty} \, \frac{|a_{n+1}|}{|a_n|} \leq \liminf_{n\to\infty} \, |a_n|^{\frac{1}{n}}$$
I've tried a number of approaches, including stuff with the AM-GM inequality, but haven't quite been able to get it out. Thanks for the help.

You may as well assume that $I = \liminf \frac{|a_{n+1}|}{|a_n|}$ is positive.
Let $0 < \epsilon < I$.  There exists an index $N$ with the property that $k \ge N$ implies $|a_{k+1}| \ge (I-\epsilon) |a_k|$.  By induction we deduce $|a_{N+k}| \ge (I-\epsilon)^k |a_N|$ for all $k \ge 0$.  Writing $n = N + k$ we have
$$
n \ge N \implies |a_n|^{1/n} \ge (I-\epsilon) \left( \frac{|a_N|}{(I-\epsilon)^N} \right)^{1/n}.
$$
The fraction in parentheses is positive and independent of $n$.  Let $n \to \infty$ to obtain
$$
\liminf_{n \to \infty} |a_n|^{1/n} \ge I-\epsilon.
$$
Now let $\epsilon \to 0^+$.
