Prove the monotoniticy of a function $f$ which satisfies $f(x) e^{f(x)} =x$

Question

Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function satisfying

$f(x) e^{f(x)} =x$ for every $x \geq 0$. Prove that $f(x)$ is monotone increasing.

I thought about taking the derivative of both sides, but I am not able to come up with a solution.

Answer 1

Let $g(u) = u e^u$. Then $g$ is increasing on $[0, \infty)$. Indeed, $g'(u) = e^u(1+u)$ is clearly positive for all $u \ge 0$.

A second observation is that $f(x) = x e^{-f(x)} \ge 0$ for all $x \ge 0$.

Now suppose that $f$ weren't increasing. Then there would exist $x, y \in [0, \infty)$ such that $x < y$ but $f(x) > f(y)$. Since $g$ is increasing on $[0, \infty)$ and $f(x), f(y) \ge 0$, then $g(f(x)) > g(f(y))$. But by hypothesis, we have $g(f(x)) = x$ and $g(f(y)) = y$, and so we get $x > y$, which is absurd. It follows that $f$ is in fact increasing.

Note: This proof doesn't even assume that $f$ is differentiable, or continuous or anything.

Answer 2

The function

$$g(t)=te^t$$ is monotonic and continuous for $t\ge0$, and its codomain is $\mathbb R^+$.

Then $g$ is invertible and monotonic in $\mathbb R^+$.