The problem is:
Suppose f(x) is a strictly concave function. Prove that x=o(f^(-1)(x)).
Do you have any suggestions on how to prove this?
Do you mean $\displaystyle x=o(f^{-1}(x))$ as $\displaystyle x\to\infty$, or as $\displaystyle x\to0$, or something else?
In any case, I don't think that this result can be true without some additional conditions. For example, the function $\displaystyle f(x) = \sqrt{x^2-1}$ is strictly concave on the interval $\displaystyle [1,\infty)$, but $\displaystyle f^{-1}(x)\approx x$ as $\displaystyle x\to\infty.$
Sorry, I didn't copy the exercise properly from the blackboard. It should have been:
Suppose $\displaystyle f(x)$ is strongly concave. Then, $\displaystyle x=o(f^{-1}(x))$ as $\displaystyle x\rightarrow \infty$.
The function you mention is strictly concave, but nog strongly concave. Thank you anyway.