Is the statement about convergence true? If for any positive integer $m,n$ there is $0 \leq x_{m+n} \leq x_m x_n$
Does $\left\{\sqrt[n]{x_{n}}\right\}$ converge? 
 A: As  said by @Ewan Delanoy  the inequality can be transformed to 
$$a_{n+m} \le a_n+a_m$$
by taking the logarithm and setting $a_n=\log x_n$.
I am following Exercise I/98 from the book of Polya/Szego (  the 4th German edition, "Aufgaben und Lehrsatze aus der Analysis I",  Springer Verlag Berlin Heideberg New York, 1970) referenced in Subadditivity
Then the sequence $\frac{a_n}{n}$ converges to the infimum
 of $\frac{a_n}{n}$ if it exists or to $-\infty$ if there is no lower bound.
This follows from
$$ \frac{a_n}{n} \le \frac{n a_1}{n}$$
and 
$$ \frac{a_{2n}}{2n} \le \frac{2a_n}{2n}$$
We set 
$$\alpha=\lim \inf \frac{a_n}{n}$$
$\alpha$ is  finite or  $-\infty$ and 
$$\frac{a_n}{n} \ge  \alpha$$
Then for $\epsilon > 0$ 
there is an $m \in \mathbb N$ such that 
$$ \alpha <= \frac{a_m}{m}<=\alpha +\epsilon $$
Each $n  \in \mathbb N$ can be written as 
$$n=qm+r, \; r \in {0,\ldots,m-1}$$
We define $a_0=0$ then
$$ \frac{a_n}{n} = \frac{a_{qm+r}}{qm+r} \lt \frac{qa_m+a_r}{qm+r}=\frac{a_m}{m}\frac{qm}{qm+r}+\frac{a_r}{n}$$
$$\alpha \le \frac{a_n}{n}  \le (\alpha +\epsilon) \frac{qm}{qm+r}+ \epsilon<=2 \epsilon, \; 
\forall n \ge \frac{\max\{a_1,\ldots,a_{r-1}\}}{\epsilon} $$
A: Answer : YES.
Hints :
1. Show we can assume wlog that $x_n > 0$ for all $n$.
2. Then we can consider $y_n={\sf log}(x_n)$. Show that
$y_n$ is subadditive.
We may then show that $(\frac{y_n}{n})$ tends to $l={\sf inf}(\frac{y_n}{n})$.
(note that $l$ will either be a real number or $-\infty$).
For example, when $l$ is finite, for any $\varepsilon \gt 0$ we may find
a $n_0$ such that $\frac{y_{n_0}}{n_0} \leq l+\varepsilon$. Using subadditivity, deduce that $\frac{y_n}{n} \leq l+2\varepsilon$ for large enough $n$.
