Example 4 (2005 Turkish Mathematical Olympiad) Find all functions $f:[0,+\infty) \rightarrow[0,+\infty)$, such that for all $x \in[0,+\infty)$, we have $4 f(x) \geqslant 3 x$, and $f[4 f(x)-3 x]=x$.


After verification, $f(x)=x$ is a function that meets the requirements of the problem.
Below, we use proof by contradiction to show that $f(x)=x$ is the unique solution.
Assume $f(x)$ satisfies the conditions and there exists $a \in [0,+\infty)$ such that $f(a) \neq a$.
Let $a_{0}=f(a), a_{1}=a, a_{n+2}=4 a_{n}-3 a_{n+1}, n=0,1, \cdots$.
From the problem statement, we know that $f\left(a_{n+1}\right)=a_{n}$, and $a_{n} \geqslant 0$.
Notice that the characteristic equation of the sequence $\left\{a_{n}\right\}$ is $t^{2}+3 t-4=0$, with roots 1 and -4, so we can set $a_{n}=c_{1} \cdot 1^{n}+c_{2} \cdot(-4)^{n}$.
Substituting $a_{0}=f(a), a_{1}=a$ yields
$$
c_{1}=\frac{4 f(a)+a}{5}, c_{2}=\frac{f(a)-a}{5} \neq 0 .
$$

If $c_{2}>0, a_{2 k+1}=c_{1}-c_{2} \cdot 4^{2 k+1}$, then
$$
\lim _{k \rightarrow+\infty} \frac{a_{2 k+1}}{4^{2 k+1}}=\lim _{k \rightarrow+\infty}\left(\frac{c_{1}}{4^{2 k+1}}-c_{2}\right)=-c_{2}<0 .
$$

Therefore, there must exist a $k$ such that $\frac{a_{2 k+1}}{4^{2 k+1}}<0$, which contradicts $a_{n} \geqslant 0$. If $c_{2}<0$, similarly, by $\lim _{k \rightarrow+\infty} \frac{a_{2 k}}{4^{2 k}}=c_{2}<0$ we derive a contradiction.
In conclusion, the only function that meets the problem's requirements is $f(x)=x$.