Hi it's a follow up of How to prove that the functional equation $f(x)+f(y)=f(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)})+f(\frac{x+y}2)$ is verified only by some basic functions? :
Conjecture :
Let $a,x>0$ and $f(x)$ be continuous, and twice differentiable on $I=(a,b)$ such that $f(x)$ is concave increasing and $f(0)=0$, $x\geq\frac{x+a}{2}\geq \frac{xf'\left(x\right)+af'\left(a\right)}{f'\left(x\right)+f'\left(a\right)}\geq a\geq \varepsilon\geq 0$ and $f'\left(x\right)+f'\left(a\right)\neq 0$ and $\frac{xf'\left(x\right)+af'\left(a\right)}{f'\left(x\right)+f'\left(a\right)}-\frac{x+a}{2}+\epsilon=0$ then it seems we have :
$$g\left(x\right)=f\left(x\right)+f\left(a\right)-f\left(\frac{x+a}{2}\right)-f\left(\frac{\left(xf'\left(x\right)+af'\left(a\right)\right)}{f'\left(x\right)+f'\left(a\right)}\right)-f(\epsilon)\le 0$$
Proof :
As we assume :
$$a\le x$$
So we have :
$$\frac{a+x}{2}-a\leq 0$$
And :
$$\frac{a+x}{2}+\frac{\left(af'\left(a\right)+xf'\left(x\right)\right)}{f'\left(a\right)+f'\left(x\right)}-a-x\le 0$$
From the constraint .And the last step follows again from one of the constraint
So applying Karamata's inequality to the function $j(x)=-f(x)$ gives the result .
Is it correct ? Have you an alternative proof ?
