I would like to prove the following statement, in generic form, for $f>0$ - concave function:
$$Af\left( \frac{x}{A}\right) > a_1f\left( \frac{x}{a_1}\right) + a_2f\left( \frac{x}{a_2}\right)+ \dots+ a_Nf\left( \frac{x}{a_N}\right), $$ where $x>0$ and: $$\sum_{i=1}^N a_i = A.$$
For example, you are trying to prove that $f(x)\gt \frac{1}{2}f(2x)+\frac{1}{3}f(3x)+\frac{1}{6}f(6x)$, with $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$.
I am not convinced this can be proven at all. Take $f(x)=\frac{1}{1-x}+C$ (with a big enough $C$), which is concave on $(1,+\infty)$, and let $x\to 1+0$. The LHS diverges (to $-\infty$) while the RHS converges (to $-\frac{3}{10}$) so the inequality must be broken for $x$ close enough to $1$.
Start with $C=0$ and then, once such $x$ is found so that the inequality is broken, increase $C$ to get to another counterexample, where the function is also positive on $[x, \infty)$.