Prove that for any $n$, we have $\sum_{i=1}^n \frac{a_i}i \ge a_n $. Consider the sequence $(a_n)_{n \ge1}$ of real nos such that $$a_{m+n}\le a_m+a_n,\ \forall m,n \ge 1$$
Prove that for any $n$, we have $$\sum_{i=1}^n \frac{a_i}i \ge a_n .$$

This is Problem 2 from the Asian Pacific Mathematical Olympiad (APMO) 1999.
I took the following proof essentially from AoPS:
We will prove this by strong induction. Note that the inequality holds for $ n=1$. Assume that the inequality holds for $ n=1,2,\ldots,k$, that is,
$$ a_1\ge a_1,\quad  a_1+\frac{a_2}{2}\ge a_2,\quad  a_1+\frac{a_2}2+\frac{a_3}3\ge a_3,\quad  \dotsc,\quad  a_1+\frac{a_2}3+\frac{a_3}3+\cdots+\frac{a_k}k\ge a_k.
$$
Sum them up:
$$ ka_1+(k-1)\frac{a_2}2a_2+\cdots+\frac{a_k}{k}\ge a_1+a_2+\cdots+a_k.$$
Add $ a_1+\ldots+a_k$ to both sides:
$$ (k+1)\left(a_1+\frac{a_2}2+\cdots+\frac{a_k}k\right)\ge (a_1+a_k)+(a_2+a_{k-1})+\cdots+(a_k+a_1)\ge ka_{k+1}.$$
Divide both sides by $ k+1$:
$$ a_1+\frac{a_2}2+\cdots+\frac{a_k}k\ge\frac{ka_{k+1}}{k+1},$$
i.e.
$$a_1 + \frac{a_2}{2} + \frac{a_3}{3} + \cdots + \frac{a_{k+1}}{k+1} \geq a_{k+1}$$
which proves the induction step.
