For any concave function we have:
$$\frac{f(0)+f(x)}{2} \leq f\left(\frac{0+x}{2}\right) \Leftrightarrow$$ for any $f(0) \geq 0$, it follows: $$f(x) \leq 2f\left(\frac{x}{2} \right).$$ How can I prove this in more generic form, i.e.: $$f(x) \leq 2f \left(\frac{x}{2} \right) \leq 3f\left(\frac{x}{3} \right) \leq \dots \leq Nf \left(\frac{x}{N} \right)$$
If $f$ is concave with $f(0) \ge 0$ and $0 < a \le b$ then $$ f(a) \ge \frac{b-a}{b-0} \, f(0) + \frac{a-0}{b-0} \, f(b) \ge \frac{a}{b} \, f(b) \\ \Longrightarrow \frac{f(a)}{a} \ge \frac{f(b)}{b} $$ Setting $a = x/v$ and $ b = x/u$ gives $$ u \, f(\frac xu) \le v \, f(\frac xv) $$ for $0 < u \le v$ and $x > 0$.