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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.

Mohsen Shahriari
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Andreas Ch.
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    The surjective part is obvious so combining it with the fact that f is strictly increasing we get bijectivity. Thus f has a unique root. If that root s not 0 from the relation above we get that $x/4$ is also a root which is impossible as f is bijective. – Andreas Ch. Jun 28 '17 at 14:11
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    Hint: Fix $a$ and suppose that $f(a)>a$. Then $\frac{a+3f(a)}{4}>a$. But $f$ is strictly increasing, so... – Kelenner Jun 28 '17 at 14:26
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    You can use the same trick to prove that the only linear map from R \to R is identity –  Jun 28 '17 at 14:31

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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.