Let $\{a_n\}$ be a sequence satisfying $\forall m,n\in\Bbb{N}:0\leq a_{m+n}\leq a_m+a_n$. Prove that $$\lim_{n\to\infty}\frac{a_n}{n}$$ exists.
At first I tried to do it with some $\epsilon-\delta$. We wish to prove that there exists some $L$ such that $\forall\epsilon>0:\exists n_0\in \Bbb{N}:n>n_0\Rightarrow|\frac{a_n}{n}-L|<\epsilon$. Which then got me to something like this:$$-\epsilon\cdot n<a_n-Ln<\epsilon\cdot n$$ which got me nowhere and I didn't find a way to make use of the inequality given.
The sequence is obviously bounded below but non-decreasing, which tells me nothing. I don't see any way to use this fact to prove the existence of given limit. Any useful tips, hints appreciated.
Partial solution: let $L = \inf \{\tfrac{a_n}n\}$ and let $\epsilon >0$. By definition of infimum, there is $k$ such that $\tfrac{a_k}k < L + \epsilon$. By induction, we find $a_{nk} \le na_k$ for all $n$. Thus $$L\le \frac{a_{nk}}{nk} \le \frac{a_k}k< L +\epsilon.$$ If you could replace $nk$ with $m$ for all $m$ large enough, you’d be done. I’ll leave this to you.
Hint: any $m$ large enough can be decomposed as $m = nk+r$ for some $n,r$ with $0\le r < k$. Take $n$ large enough and use $a_{nk+r} \le a_{nk} + a_r$.
EDIT: full solution. With $L, \epsilon, k$ as defined above, take $M$ large enough that $$\frac{\max\{a_1, a_2, \ldots, a_{k-1}\}}{M} < \epsilon.$$
Then for any $m > Mk$, there is $N\ge M$ and $ 0 \le r < k$ such that $m = Nk+r$. Now $$L \le \frac{a_m}{m} = \frac{a_{Nk+r}}{Nk+r} \le \frac{a_{Nk}}{Nk+r} + \frac{a_r}{Nk+r} \le \frac{a_{Nk}}{Nk} + \frac{a_r}{M} \le \frac{a_k}{k} + \epsilon < L+2\epsilon.$$ This shows that for any $m > Mk$, we have $\lvert \tfrac{a_m}{m} - L \rvert < 2\epsilon$. Hence by the definition of limit, $\tfrac{a_m}{m} \to L$.