Is $f(2x)/f(x)$ nonincreasing for concave functions with $f(0)=0$?

I have a question about concave functions.

Let $f:R_+\rightarrow R_+$ be any nonidentically zero, nondecreasing, continuous, concave function with $f(0)=0$. Do we have that the ratio function $f(2x)/f(x)$ is nonincreasing on $(0,+\infty)$?

No. Let $f$ be the function $$ f(x) \;=\; \begin{cases}2x & \text{if }0\leq x \leq 1 \\ x+1 & \text{if }1 \leq x.\end{cases} $$ Then $f$ satisfies the hypotheses you have given, but $$ \frac{f(2)}{f(1)} = \frac{3}{2} \qquad\text{and}\qquad \frac{f(4)}{f(2)} = \frac{5}{3} > \frac{3}{2}\text{,} $$ so $f(2x)/f(x)$ increases from $x=1$ to $x=2$.

Edit: There are also differentiable counterexamples, e.g. $f(x) = (x^2+10x)/(x+1)$.