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: $$\begin{array}{rl}f(a_1+a_2+\cdots +a_{2^{k+1}})& =f((a_1+a_2+\cdots +a_{2^k})+(a_{2^k+1}+\cdots +a_{2^{k+1}}))\\ & \le 2\max \{f(a_1+a_2+\cdots +a_{2^k}),f(a_{2^k+1}+\cdots +a_{2^{k+1}})\}\\ & \le 2\max \{2^k\max \{f(a_1),f(a_2),\dots ,f(a_{2^k})\},2^k\max \{f(a_{2^k+1}),\dots ,f(a_{2^{k+1}})\}\}\\ & \le 2^{k+1}\max \{f(a_1),f(a_2),\dots ,f(a_{2^{k+1}})\}\end{array}$$ Now suppose that $2^{k-1}<n\le 2^k$: $$\begin{array}{rl}f(a_1+\cdots +a_n)& =f(a_1+\cdots +a_n+\underset{2^k-n}{\underbrace{0+0+\cdots +0}})\\ & \le 2^k\max \{f(a_1),\dots ,f(a_n),f(0),\dots ,f(0)\}\\ & =2^k\max \{f(a_1),\dots ,f(a_n)\}\le 2n\max \{f(a_1),\dots ,f(a_n)\}\end{array}$$ If we put $a_1=a_2=\cdots =a_n=1$ then $f(n)=f(\underset{n}{\underbrace{1+1+\cdots +1}})\le 2nf(1)$. Therefore $$\begin{array}{rl}(f(a+b){)}^n& =f((a+b{)}^n)=f\left(\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)a^i{b}^{n-i}\right)\le 2(n+1)\underset{0\le i\le n}{\max }\{f\left(\left(\genfrac{}{}{0ex}{}{n}{i}\right)a^i{b}^{n-i}\right)\}\\ & \le 2(n+1)\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)f(a^i)f(b^{n-i})\le 4(n+1)f(1)\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)f(a{)}^{i}f(b{)}^{n-i}\\ & =4(n+1)f(1)(f(a)+f(b){)}^n\Rightarrow f(a+b)\le \sqrt[n]{4(n+1)f(1)(f(a)+f(b))}\end{array}$$ If $n$ tends to infinity, we have $\sqrt[n]{4(n+1)f(1)}\to 1$ and this implies the desired result. $\square$