Example 1 Given that $f(x)$ is a polynomial function, solve the functional inequality
$$
f(x+1)+f(x-1) \leqslant 2 x^{2}-4 x .
$$
Given that $f(x)$ is a polynomial function, and $f(x+1)$ and $f(x-1)$ do not change the degree of $f(x)$, it follows from (1) that the degree of $f(x)$ is at most 2. Let's assume $f(x) = a x^{2} + b x + c$, then
$$
\begin{array}{l}
f(x+1) = a(x+1)^{2} + b(x+1) + c = a x^{2} + (2 a + b) x + (a + b + c), \\
f(x-1) = a(x-1)^{2} + b(x-1) + c = a x^{2} + (b - 2 a) x + (a - b + c),
\end{array}
$$

The original inequality becomes
$$
f(x+1) + f(x-1) = 2 a x^{2} + 2 b x + 2(a + c) \leqslant 2 x^{2} - 4 x,
$$

which simplifies to
$$
\begin{array}{c}
(a-1) x^{2} + (b+2) x + (a+c) \leqslant 0, \\
\left\{\begin{array}{l}
a - 1 = 0, \\
b + 2 = 0, \\
a + c \leqslant 0,
\end{array} \Rightarrow \left\{\begin{array}{l}
a = 1, \\
b = -2, \\
c \leqslant -1,
\end{array}\right.\right.
\end{array}
$$

Therefore,
$$
f(x) = x^{2} - 2 x + c \quad (c \leqslant -1),
$$

or
$$
\left\{\begin{array}{l}
a - 1 < 0, \\
\Delta = (b + 2)^{2} - 4(a - 1)(a + c) \leqslant 0,
\end{array}\right.
$$
which simplifies to
$$
\left\{\begin{array}{l}
a < 1, \\
c \leqslant \frac{(b + 2)^{2}}{4(1 - a)} - a.
\end{array}\right.
$$
Thus, the solution to the given inequality is $f(x) = x^{2} - 2 x + c \quad (c \leqslant -1)$ or $f(x) = a x^{2} + b x + c$, satisfying
$$
\left\{\begin{array}{l}
a < 1, \\
c \leqslant \frac{(b + 2)^{2}}{4(1 - a)} - a.
\end{array}\right.
$$

Verification: It is easy to verify that the above solutions indeed satisfy (1).