Five. (Full marks 20 points) Given the function $f(x)=a x^{2} + b x + c$ whose graph 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.

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Five, assuming there exist $a, b, c$ that meet the conditions.
$\because f(x)$'s graph passes through $(-1,0)$,
$\therefore f(-1)=0$, i.e., $a-b+c=0$.
Also, $\because x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$, let $x=1$, then
$$
\begin{array}{l}
1 \leqslant a+b+c \leqslant \frac{1}{2}\left(1+1^{2}\right)=1 . \\
\therefore a+b+c=1 . \\
\therefore b=\frac{1}{2}, a+c=\frac{1}{2} . \\
\therefore f(x)=a x^{2}+\frac{1}{2} x+\left(\frac{1}{2}-a\right) . \\
\text { By }\left\{\begin{array}{l}
f(x) \geqslant x, \\
f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right) \\
\left(a-\frac{1}{2}\right) x^{2}+\frac{1}{2} x-a \leqslant 0 .
\end{array}\right. \\
\left\{\begin{array}{l}
a x^{2}-\frac{1}{2} x+\left(\frac{1}{2}-a\right) \geqslant 0, \\
(
\end{array}\right.
\end{array}
$$

According to the problem, for any real number $x$, (1) and (2) both hold.
For (1), if $a=0$, then $x \leqslant 1$, which does not meet the problem's requirements; if $a>0$,
to make the solution set of (1) $\mathbf{R}$, we need
$$
\left\{\begin{array} { l } 
{ a > 0 , } \\
{ \Delta \leqslant 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
a>0, \\
\frac{1}{4}-4 a\left(\frac{1}{2}-a\right) \leqslant 0 .
\end{array}\right.\right.
$$

Solving it, we get $a=\frac{1}{4}$.
For $a=\frac{1}{4}$, consider (2) again. Substituting $a=\frac{1}{4}$ into (2) gives
$$
x^{2}-2 x+1 \geqslant 0 \text {. }
$$

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