Let $f: \mathbb{R}_{+} \rightarrow \mathbb{R}$ be a strictly monotonic function.
What kind of extra conditions we need to add, in order to ensure that this function $f$ and its inverse function $f^{-1}$ are both subadditive?
From my own opinion, when we are trying to think about this question, it might be convenient to relax the domain of $f$ first, such that $f$ maps $\mathbb{R}$ to $\mathbb{R}$.
Any ideas are most welcome! Thanks in advance^-^
Suppose there is such a function $f$ that is strictly increasing, and both $f$ and its inverse are subadditive. Note then that its inverse $f^{-1}$ must be strictly increasing.
Assume there are $x, y$ so that we have $f(x+y) < f(x) + f(y)$. Then
$$ f^{-1} (f(x+y)) < f^{-1}(f(x)+f(y)) \leq f^{-1}(f(x)) + f^{-1}(f(y))$$
or $x + y < x+y$, which is a contradiction. This shows that any such function $f$ must have the property $f(x+y) = f(x) + f(y)$, for all $x, y$.
It's not too hard to see that for any integer $n$ this means that $f(nx) = n f(x)$, and it's not too hard to then extend this result for any rational number $q$. Thus, $f(q) = f(1) q = c q$. Since $f$ is increasing, we can assume $c = f(1) > 0$.
Suppose $f(x) \neq cx$ for some real $x$. Then we have some rational $q$ such that (wlog) $$ x < q < \frac{f(x)}{c}$$
so that $f(x) < f(q) = cq$. But then $\frac{f(x)}{c} < q$ which is a contradiction.
Thus, the only solutions must be of the form $f(x) = cx$ where $c > 0$.