Iranian Mathematical Olympiad

We claim that for every $k \in \mathbb{N}$ and real numbers $a_1, a_2, \dots, a_{2^k}$:
$$f(a_1 + a_2 + \cdots + a_{2^k}) \le 2^k \max\{f(a_1), f(a_2), \dots, f(a_{2^k})\}$$
Proof is done by induction on $k$. Basis is obviously the condition (iii). Suppose the claim is true for $k$. For $k+1$ we have:
$$\begin{align*} f(a_1 + a_2 + \cdots + a_{2^{k+1}}) &= f((a_1 + a_2 + \cdots + a_{2^k}) + (a_{2^k+1} + \cdots + a_{2^{k+1}})) \\ &\le 2 \max\{f(a_1 + a_2 + \cdots + a_{2^k}), f(a_{2^k+1} + \cdots + a_{2^{k+1}})\} \\ &\le 2 \max\{2^k \max\{f(a_1), f(a_2), \dots, f(a_{2^k})\}, 2^k \max\{f(a_{2^k+1}), \dots, f(a_{2^{k+1}})\}\} \\ &\le 2^{k+1} \max\{f(a_1), f(a_2), \dots, f(a_{2^{k+1}})\} \end{align*}$$
Now suppose that $2^{k-1} < n \le 2^k$:
$$\begin{align*} f(a_1 + \cdots + a_n) &= f(a_1 + \cdots + a_n + \underbrace{0 + 0 + \cdots + 0}_{2^k - n}) \\ &\le 2^k \max\{f(a_1), \dots, f(a_n), f(0), \dots, f(0)\} \\ &= 2^k \max\{f(a_1), \dots, f(a_n)\} \le 2n \max\{f(a_1), \dots, f(a_n)\} \end{align*}$$
If we put $a_1 = a_2 = \cdots = a_n = 1$ then $f(n) = f(\underbrace{1+1+\cdots+1}_{n}) \le 2nf(1)$. Therefore
$$\begin{align*} (f(a+b))^n &= f((a+b)^n) = f\left(\sum_{i=0}^{n} \binom{n}{i} a^i b^{n-i}\right) \le 2(n+1) \max_{0 \le i \le n} \{f\left(\binom{n}{i} a^i b^{n-i}\right)\} \\ &\le 2(n+1) \sum_{i=0}^{n} \binom{n}{i} f(a^i) f(b^{n-i}) \le 4(n+1)f(1) \sum_{i=0}^{n} \binom{n}{i} f(a)^i f(b)^{n-i} \\ &= 4(n+1)f(1)(f(a)+f(b))^n \Rightarrow f(a+b) \le \sqrt[n]{4(n+1)f(1)(f(a)+f(b))} \end{align*}$$
If $n$ tends to infinity, we have $\sqrt[n]{4(n+1)f(1)} \to 1$ and this implies the desired result. $\square$