Prove an inequality with Jensen's inequality and its companion Problem :
Let $f(x)$ be continuous convex  increasing on $(0,\infty)$ such that $f(\alpha)=0$ ; $x,a,\alpha>0$ with the constraint :
$$\infty>\alpha\geq x\geq \frac{xf'\left(x\right)+af'\left(a\right)}{f'\left(x\right)+f'\left(a\right)}\geq \frac{\left(x+a\right)}{2}\geq a $$
Where $\forall n\ge 2$ :$$f(x)=\sum_{i=0}^{n}a_ix^i$$ And $a_0 \ge a_1 \ge ... \ge a_n>0$
Then we have for $a,x,\alpha$ sufficiently large :
$$f\left(\frac{\left(x+a\right)}{2}\right)+f\left(\frac{\left(xf'\left(x\right)+af'\left(a\right)\right)}{f'\left(x\right)+f'\left(a\right)}\right)-f\left(a\right)-f\left(x\right)+f(\alpha)\geq f(\alpha)$$


My attempt :
We use majorization (with an alternative criteria see Aops page) as first step:
$$\alpha=\alpha$$
And :
$$\alpha+x\geq \alpha+\frac{xf'\left(x\right)+af'\left(a\right)}{f'\left(x\right)+f'\left(a\right)}$$
Finally :
$$\frac{x+a}{2}+\frac{xf'\left(x\right)+af'\left(a\right)}{f'\left(x\right)+f'\left(a\right)}-a-x\leq 0$$
Wich is wrong .


How to prove it ?
 A: I take a simple example :
Let $x>0$ such that $f(x)=x+kx^2$ with $0<k<1$ and $a=0$ , $b>0$ :
Then it seems we have the equality :
$$f\left(\frac{\left(x+a\right)}{2}\right)+f\left(\frac{\left(xf'\left(x\right)+af'\left(a\right)\right)}{f'\left(x\right)+f'\left(a\right)}\right)-f\left(a\right)-f\left(x\right)=\frac{\left(f\left(a\right)-f\left(b\right)\right)}{a-b}\left(\frac{\left(x-a\right)}{2}\right)+f\left(a\right)-f'\left(x\right)\left(x-\frac{\left(xf'\left(x\right)+af'\left(a\right)\right)}{f'\left(x\right)+f'\left(a\right)}\right)+\frac{f''\left(x\right)}{2}\left(\left(x-\frac{\left(xf'\left(x\right)+af'\left(a\right)\right)}{f'\left(x\right)+f'\left(a\right)}\right)\right)^{2}$$
For $x=2b$
Perhaps we can generalize this example but it seems really hard .
Edit 27/06/2022 :
We have a simple generalization via this example :
Let $m=2,a=k=0.5,b=0.9,x\geq 1/8$
Now and very simply we add an $x$ on :
$$r\left(x\right)=x\left(f\left(\frac{a+x}{2}\right)+f\left(\frac{\left(xf'\left(x\right)+af'\left(a\right)\right)}{f'\left(x\right)+f'\left(a\right)}\right)-f\left(x\right)-f\left(a\right)\right)$$
Where :
$$f(x)=\left(2+x+kx^{2}\right)$$
And now for $m=2,a=k=0.5,b=0.9,x\geq 1/8$ it seems we have the inequality :
$$r(x)\leq p\left(x\right)=g\left(\frac{a+x}{2}\right)+g\left(\frac{\left(xg'\left(x\right)+ag'\left(a\right)\right)}{g'\left(x\right)+g'\left(a\right)}\right)-g\left(x\right)-g\left(a\right)$$
Where $g\left(x\right)=x\left(m+x+kx^{2}\right)$
