Concavity of $f(z)$ for $z>1$ given $f(\frac xy)>\frac xyf'(\frac xy)$ Edit: Had typo in first line. Should've been that $f'(x)>0$ for $z>1$
Assume 


*

*$f'(z)>0$ for $z>1$

*$f'(z) <0$ for $z<1$, and

*$f(z) = f\left(\dfrac{1}{z}\right)$.


*

*(if knowing a domain helps, you can assume the domain of $f$ is $0$ to some bound $B>1$. 



My question is:

Does $f\left(\dfrac{x}{y}\right)>\dfrac{x}{y}f'\left(\dfrac{x}{y}\right)$ for all $x>y$ imply $f(z)$ is concave for all $z>1$?

Any insight into this question? I am not quite sure how to approach it (or if it is true).
My thought was to try getting to something like $$f'(z)(x-z) \geq f(x)-f(z),$$ but I did not succeed.
 A: $\def\e{\mathrm{e}}$This is not true. Consider$$
f(x) = \exp\left( -\frac{1}{x - 1} \right) + 1. \quad \forall x > 1
$$
First, for $x > 1$,$$
f'(x) = \frac{1}{(x - 1)^2} \exp\left( -\frac{1}{x - 1} \right),\quad f''(x) = -\frac{2x - 3}{(x - 1)^4} \exp\left( -\frac{1}{x - 1} \right),
$$
thus $f$ is neither concave nor convex on $(1, +\infty)$. Now define$$
F(x) = f(x) - x f'(x). \quad \forall x > 1
$$
Because$$
F'(x) = -x f''(x) = \frac{(2x - 3)x}{(x - 1)^4} \exp\left( -\frac{1}{x - 1} \right),
$$
then for any $x > 1$,$$
F(x) \geqslant F\left( \frac{3}{2} \right) = 1 - 5\e^{-2} > 0,
$$
which implies$$
f\left( \frac{x}{y} \right) > \frac{x}{y} f'\left( \frac{x}{y} \right). \quad \forall x > y > 0
$$

If further defining$$
f(x) = \begin{cases}
f\left( \dfrac{1}{x} \right); & 0 < x < 1\\
1; & x = 1
\end{cases},
$$
it can be proved that $f \in C^\infty(\mathbb{R}_+)$. Now, $f(x) = f\left(\dfrac{1}{x}\right)$ obviously holds for $x > 0$ and$$
f'(x) = \begin{cases}
-\dfrac{1}{(1 - x)^2} \exp\left( -\dfrac{x}{1 - x} \right); & 0 < x < 1\\
0; & x = 1\\
\dfrac{1}{(x - 1)^2} \exp\left( -\dfrac{1}{x - 1} \right); & x > 1
\end{cases}
$$
