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.
 A: 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 $\epsilon \to 0^+$.
A: Assuming $a_0 = 1$, $a_n > 0$ for all $n$.
Set $b_n = a_{n+1} / a_{n}$, then $a_n = \prod_{i=0}^{n-1}{b_i}$.  So this is equivalent to the original expression:
$$\liminf_{n\to\infty} \, b_n \leq \liminf_{n\to\infty} \, (\prod_{i=0}^n b_i)^{\frac{1}{n}}$$
Take the log of both sides:
$$\liminf_{n\to\infty} \, c_n \leq \liminf_{n\to\infty} \, {\frac{1}{n}}(\sum_{i=0}^n c_i)$$
which is implied by "the average of a set of elements is greater than or equal to the smallest element in the set".
