Bounded sequence of reals Suppose that $\{a_{n} \}$ is a sequence of real numbers that satisfy $a_{i} + a_{j} \geq a_{i+j}$. 
Then prove $$a_{n} \leq \sum_{i=1}^{n} \frac{a_{i}}{i}$$. 
I tried to use straight-up induction, but I am getting stuck. What I have is:
$a_{n+1} \leq a_{1} + a_{n} \leq a_{1} + \sum_{i=1}^{n} \frac{a_{i}}{i}$, but I don't seem to be getting anywhere.

From the inequality you easily obtain that for all integers $i_1, ..., i_k$, one has: $a_{i_1} + ... + a_{i_k} \geq a_{i_1 + ... + i_k}$. By taking them all equal, one gets: $a_{kl} \leq l \cdot a_k$, so $\frac{a_{kl}}{l} \leq a_k$.
We have:
$$
a_1 + a_{n-1} \geq a_n \\
... \\
a_{n-1} + a_{1} \geq a_n \\
$$
Summing all the inequalities gives $(n-1) \cdot a_n \leq 2\cdot(a_1 + \cdots + a_{n-1})$. Let us use this to prove the induction.
Assume the inequality is true for integers $k \leq n$. We will show it stands for $n+1$.
From what we proved we get $n \cdot a_{n+1} \leq 2\cdot(a_1 + \cdots + a_{n})$.
Thus,
$$
\begin{align*}
n \cdot a_{n+1} 
&\leq 2\cdot \left( a_1 + (a_1 + \frac{a_2}{2}) + \cdots + (a_1 + \cdots + \frac{a_n}{n}) \right) \\
&\leq 2\cdot \left( n \cdot a_1 + (n-1) \cdot \frac{a_2}{2} + \cdots + 1 \cdot \frac{a_n}{n} \right) \\
\end{align*}
$$
Adding $a_{n+1}$ on both sides gives:
$$
\begin{align*}
(n+1) \cdot a_{n+1} 
&\leq 2\cdot \left( n \cdot a_1 + (n-1) \cdot \frac{a_2}{2} + \cdots + 1 \cdot \frac{a_n}{n} \right) + a_{n+1}  \\
\end{align*}
$$
But I do not know how to go further...
