Solving the functional equation $f\big(4f(x)-3x\big)=x$ Find all functions $f:[0,+\infty)\to [0,+\infty)$ such that $f(x)\geq \frac{3x}{4}$ and
$$f\big(4f(x)-3x\big)=x,\forall x\in[0,+\infty)$$
 A: Note that $4f(x) - 3x\geq 0$ for all $x\geq 0$.
Given a lower bound of the form $f(x) \geq a x$ with $a>0$ for $x \geq 0$. Let $x > 0$ and set $y = 4f(x) - 3x \geq 0$. Then
  $$f(y) = x \geq a y = a(4f(x) - 3x) = 4af(x) - 3 ax,$$
so $f(x) \leq \frac{(3a + 1)x}{4a}$.
Similarly, given an upper bound of the form $f(x)\leq ax$ with $a>0$ for $x \geq 0$, we may again let $x > 0$ and $y = 4f(x) - 3x$ to get
  $$f(y) = x \leq ay = 4af(x) - 3ax.$$
Thus, we get a lower bound $f(x) \geq \frac{(3a + 1)x}{4a}$.
Set $a_0 = \frac{3x}{4}$ and define inductively $a_{n+1} = \frac{3a_n + 1}{4a_n} = \frac{3}{4} + \frac{1}{4a_n}$ for $n > 0$. Note that $a_n > 0$ for all $n$ and that, by the above calculations, for all $x\geq 0$, $a_nx\leq f(x)\leq a_{n+1} x$ for all even $n$. We claim that
  $$\lim_{n\to \infty} a_n = 1.$$
Applying the recurrence relation twice yields
  $$a_{n + 2} = \frac{3\frac{3a_n + 1}{4a_n} + 1}{4\frac{3a_n + 1}{4a_n}} = \frac{13a_n + 3}{12a_n + 4} = 1 + \frac{a_n - 1}{a_n + 4}.$$
From this one can check that $a_{n-2}\leq a_n\leq 1$ when $n$ is even and that $a_{n-2}\geq a_n\geq 1$ when $n$ is odd. In particular, the sequences $\{a_n\mid n\text{ even}\}$ and $\{a_n\mid n\text{ odd}\}$ are bounded monotone and thus convergent. Let $a^-$, resp. $a^+$, be the limit of the even, resp. odd sequence, so that $a^-\leq 1\leq a^+$. Both $a^+$ and $a^-$ must satisfy
  $$a = 1 + \frac{a - 1}{a + 4},$$
i.e., $a^2 + 2a - 3 = (a + 2)(a - 1) = 0$. As $a^\pm$ are positive, the only possibility is $a = 1$.
As noted above, we have
  $$a_n x \leq f(x) \leq a_{n+1} x$$
for all $n$ even, so $f(x) = x$ is the only possibility.
