If $f$ is a real valued function defined on the set of real numbers and $f$ is strictly increasing on its domain and the following holds: $$f\left(\frac{x+3f(x)}{4}\right)=x$$ for all real $x$, then prove that $f(x)=x$ for all real $x$.
I've proven that $f(0)=0$ and that $f$ is bijective but I don't see anything else.
Notice that $x$ and $f(x)$ have the same sign. Now we proceed by contradiction.
Assume that for some $x \in \mathbb{R}$, $f(x) > x$; then $$\frac{x + 3f(x)}{4} > x $$ Now because $f$ is strictly increasing, this imples $$f\bigg(\frac{x+3f(x)}{4}\bigg) > f(x)$$ $$x > f(x) $$ Hence we get a contradiction.
PS: For the initial steps, read the comment of the OP to his question.