Working with Slater's inequality (a companion of Jensen's inequality) I found this statement:



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> > Let $f(x)$ be a continuous, twice differentiable function, convex or concave and non constant on $(0,\infty)$ and increasing on $(\alpha,\infty)$ with $\alpha>0$ a constant. Now define:
> > $$g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$$
> > Claim: $g(x)$ has an asymptote as $x\to \infty$
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I have not the key to approach the general case but let me try some examples:


For my first example I take the exponential ([see here](https://www.wolframalpha.com/input/?i=limit%20%20&assumption=%7B%22F%22%2C%20%22Limit%22%2C%20%22limit%22%7D%20-%3E%22infinity%22&assumption=%7B%22F%22%2C%20%22Limit%22%2C%20%22limitfunction%22%7D%20-%3E%222%20%28e%5E%282%20x%29%2Be%5E%282%29%29%2F%28e%5E%28x%29%2Be%5E%281%29%29-2%20e%5E%28%28x%20e%5E%28x%29%2Be%5E%281%29%29%2F%28e%5E%28x%29%2Be%5E%281%29%29%29%2B4e-2*e*x%22&assumption=%7B%22C%22%2C%20%22limit%22%7D%20-%3E%20%7B%22Calculator%22%7D)). The function can be decreasing or increasing on a such interval, see for example the function $f(x)=x^x$. As further particular cases we have $f(x)=\ln(x)$ or $f(x)=\arctan(x)$ and for these the asymptotes are constants. So far I have tried only elementary functions and their compositions but it would be curious that it works only for them.


**My question**:  

Do you know a counter-example (it would be perfect because I have some doubt on it) or a proof (which I think is not easy) for this statement?


Thanks in advance!


**Update**: I cannot post it again new as a new question so I edit this with a new possible theorem:



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> > Let $f(x)$ be a continuous, $n$ times differentiable, convex and non constant on $(0,\infty)$, increasing on $(1,\infty)$ and finally with unbounded derivatives function, then define:
> > $$g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$$
> > With $g(x)$ strictly increasing on $(1,\infty)$.
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Claim: $g(x)$ have an asymptote as $x\to \infty$.


[Cross posted on math.Se](https://math.stackexchange.com/questions/3920773/existence-of-an-asymptote-for-gx-fracfxfxf1f1fxf1-f-le)



Existence of an asymptote for $g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$
Here is one counter example: $f(x)=\sqrt x$ gives a function $g(x)=2-x^{1/4}-\dfrac{2}{1+\sqrt{x}}$ without an  [asymptote](https://en.wikipedia.org/wiki/Asymptote)  for $x\rightarrow\infty$.


*Note: this is a counter example to the original conjecture in the OP, not to the updated conjecture.*


