Example 4 (2000 Hebei Province Competition Question) Given that the graph of the function $f(x)=a x^{2}+b x+c$ passes through the point $(-1,0)$. Does there exist constants $a, b, c$ such that the inequality $x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$? If they exist, find the values of $a, b, c$; if not, explain the reason.
Assume there exist $a, b, c$ that meet the conditions.
Since the graph of $f(x)$ passes through $(-1,0)$,
then $f(-1)=0$, i.e., $a-b+c=0$.
Also, since $x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$, let $x=1$, then $1 \leqslant a+b+c \leqslant \frac{1}{2}\left(1+1^{2}\right)=1$.
Thus, $a+b+c=1$.
Therefore, $b=\frac{1}{2}, a+c=\frac{1}{2}$.
Hence, $f(x)=a x^{2}+\frac{1}{2} x+\left(\frac{1}{2}-a\right)$.
From $\left\{\begin{array}{l}f(x) \geqslant x, \\ f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)\end{array}\right.$,
we get $\left\{\begin{array}{l}a x^{2}-\frac{1}{2} x+\left(\frac{1}{2}-a\right) \geqslant 0, \\ \left(a-\frac{1}{2}\right) x^{2}+\frac{1}{2} x-a \leqslant 0 .\end{array}\right.$
According to the problem, for any real number $x$, (1) and (2) must both hold.
For (1), if $a=0$, then $x \leqslant 1$, which does not meet the requirement; if $a>0$, to make the solution set of (1) $\mathbf{R}$, we need $\left\{\begin{array}{l}a>0, \\ \Delta \leqslant 0,\end{array}\right.$ i.e., $\left\{\begin{array}{l}a>0, \\ \frac{1}{4}-4 a\left(\frac{1}{2}-a\right) \leqslant 0 .\end{array}\right.$
Solving it, we get $a=\frac{1}{4}$.
For $a=\frac{1}{4}$, consider (2) again. Substituting $a=\frac{1}{4}$ into (2) yields
$$
x^{2}-2 x+1 \geqslant 0 \text {. }
$$

Its solution set is $\mathbf{R}$.
Therefore, there exist $a, b, c$ that meet the conditions, where $a=b=c=\frac{1}{4}, b=\frac{1}{2}$.