Example 3 Given the functional inequality
$$
f(x+p)-q f(x) \geqslant a x+b .
$$

Try to find $f(x)$, where $p, q, a, b$ are constants, and $q \neq 1$.
The solution to the corresponding functional equation of (1) is usually obtained by the iterative method:
$$
f(x)=\frac{a x}{1-q}+c q^{\frac{1}{x}}+\frac{b}{1-q}-\frac{a p}{(1-q)^{2}}.
$$

Under certain conditions, this answer provides a particular solution to the functional inequality (1). In general, this answer may be meaningless. For example, when $p=0, c \neq 0$, the solution to the functional inequality (1) is $f(x)=\frac{a x+b}{1-q}$.
Below, we solve the functional inequality (1) using the method of case analysis.
Solution When $p=0$, from (1) we immediately get the solution to the functional inequality as
$$
f(x) \geqslant \frac{a x+b}{1-q}(q1).
$$

When $p \neq 0$, let $f(x)=g(x)+\lambda x+\mu$. Here, $g(x)$ is a function to be determined, and $\lambda, \mu$ are constants to be determined. Substituting into the functional inequality (1), we get
$$
g(x+p)+\lambda(x+p)+\mu-q[g(x)+\lambda x+\mu] \geqslant a x+b,
$$

which simplifies to $g(x+p)-q g(x)+(\lambda-\lambda p-a) x+\lambda p+\mu-q \mu-b \geqslant 0$.
Let $\lambda-\lambda p-a=0$ and $\lambda p+\mu-q \mu-b=0$, we get
$$
\lambda=\frac{a}{1-q}, \mu=\frac{b}{1-q}-\frac{a p}{(1-q)^{2}},
$$

and $g(x+p) \geqslant q g(x)$.
When $q>0$, (2) is equivalent to
$$
\frac{g(x+p)}{\left(q^{\frac{1}{\alpha}}\right)^{x+p}} \geqslant \frac{g(x)}{\left(q^{\frac{1}{\alpha}}\right)^{\prime}}, \text{ let } h(x)=\frac{g(x)}{\left(q^{\frac{1}{\alpha}}\right)^{\prime}}.
$$

Then $g(x)=h(x) q^{\frac{1}{x}}$, and the above equation becomes $h(x+p) \geqslant h(x)$. By back-substitution, the general solution to the functional inequality (1) in this case is
$$
f(x)=\frac{a x}{1-q}+h(x) q^{\frac{1}{\hbar}}+\frac{b}{1-q}-\frac{a p}{(1-q)^{2}}, x \in D.
$$

where $h(x)$ is any function defined on the real set $D(\subseteq \mathbf{R})$ that satisfies $h(x+p) \geqslant h(x)$. When $q<0$, (2) is equivalent to
$$
\frac{g(x+p)}{\left[(-q)^{\frac{1}{p}}\right]^{x+p}} \geqslant-\frac{g(x)}{\left[(-q)^{\frac{1}{p}}\right]^{\prime}}, \text{ let } h(x)=\frac{g(x)}{\left[(-q)^{\frac{1}{p}}\right]^{x}},
$$

then $g(x)=h(x)(-q)^{+}$, and the above equation becomes $h(x+p) \geqslant-h(x)$. By back-substitution, the general solution to the functional inequality (1) in this case is
$$
f(x)=\frac{a x}{1-q}+h(x)(-q)^{\frac{1}{p}}+\frac{b}{1-q}-\frac{a p}{(1-q)^{2}}, x \in D.
$$

where $h(x)$ is any function defined on the real set $D(\subseteq \mathbf{R})$ that satisfies $h(x+p) \geqslant-h(x)$.