Prove that $f(m+n) \leq \max(f(m),f(n))$ Let $f$ be a function defined on the set of positive integers such that for each $m,n \in \mathbb{Z}_{+}$,
(i) $f(mn)=f(m)f(n)$,
(ii) $f(m+n) \leq f(m)+f(n)$,
(iii) $0 \leq f(m) \leq 1$.
How can I prove that $f(m+n) \leq \max(f(m),f(n))$?
The hint says to consider $\{f(m+n)\}^k$ for each positive number $k$, but cannot find the way...

You have for any $k \ge 1$, 
$$\begin{eqnarray*}
f(m+n)^k &=& f\left((m+n)^k\right) \\
&=& f\left(\sum_{i = 0}^k \binom{k}{i} \, m^i n^{k-i} \right)\\
&\le& \sum_{i = 0}^k \underbrace{f\binom{k}{i}}_{\le 1} \ f(m)^i f(n)^{k-i} \\
&\le& \sum_{i = 0}^k \ f(m)^i f(n)^{k-i} \\
&\le& (k+1) . \max\{f(m), f(n)\}^k
\end{eqnarray*}$$
So for all $k \ge 1$, we have $f(n+m) \le (k+1)^{\frac{1}{k}} . \max\{f(m), f(n)\}$. Since $(k+1)^{\frac{1}{k}} \underset{k \to \infty}{\longrightarrow} 1$, we get
$$f(n+m) \le \max\{f(m), f(n)\}$$
To provide some context, (ii) is called the triangle inequality, and $f(n+m) \le \max\{f(m), f(n)\}$ is the ultrametric inequality.
