Example 1 Find two particular solutions of the functional inequality
$$
f(x+y) \geqslant 2008 f(x)-2009 f(y)
$$

on the set of real numbers $\mathrm{R}$ or its subsets.
If $f(x)=c$ (a constant) is a particular solution of the functional inequality (1), substituting it into (1) yields $c \geqslant 2008 c - 2009 c$,

so $c \geqslant 0$. That is, $f(x)=c(c \geqslant 0)$ is a particular solution of (1) on $\mathbf{R}$.
If $f(x)=x+\frac{1}{2}$ is another particular solution of the functional inequality, then substituting it into (1) gives
$$
x+y+\frac{1}{2} \geqslant 2008\left(x+\frac{1}{2}\right)-2009\left(y+\frac{1}{2}\right)
$$

which simplifies to $2007 x - 2010 y - 1 \leqslant 0$.
Solving the system of equations $\left\{\begin{array}{l}y=x+\frac{1}{2} \\ 2007 x-2010 y-1=0\end{array}\right.$ yields $x=-\frac{1006}{3}$.
Thus, $f(x)=x+\frac{1}{2}, x \in\left[-\frac{1006}{3},+\infty\right)$ is a particular solution of (1).
Therefore, we have found two particular solutions of the functional inequality (1):
$$
f(x) \doteq c(c \geqslant 0), f(x)=x+\frac{1}{2}, x \in\left[-\frac{1006}{3},+\infty\right) .
$$