Example 4 Let the function $y=f(x)$ have the domain $\mathbf{R}$, and when $x>0$, we have $f(x)>1$, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x) f(y)$. Solve the inequality
$$
f(x) \leqslant \frac{1}{f(x+1)} .
$$

Analysis: This problem involves an abstract function, and we can find a prototype function $f(x)=a^{x}(a>1)$. Therefore, we conjecture that $f(x)$ has the following properties: $f(x)>0, f(0)=1, f(x)$ is an increasing function on $\mathbf{R}$.
(1) Proof that $f(x)>0$ always holds.
From $f(x)=f\left(\frac{x}{2}+\frac{x}{2}\right)=f^{2}\left(\frac{x}{2}\right) \geqslant 0$, if there exists $x_{0} \in \mathbf{R}$, such that $f\left(x_{0}\right)=0$, then for any real number $x$, we have
$$
f(x)=f\left(x_{0}+x-x_{0}\right)=f\left(x_{0}\right) f\left(x-x_{0}\right)=0 .
$$

This contradicts the given condition, hence $f(x)>0$.
(2) Proof that $f(0)=1$.

In $f(x+y)=f(x) f(y)$, let $x=y=0$,
we have
$f(0)=f^{2}(0)$.
From (1) we know $f(0)>0$. Therefore, $f(0)=1$.
(3) Proof that $f(x)$ is an increasing function on $\mathbf{R}$.
Let $x_{2}>x_{1}$, then $x_{2}-x_{1}>0$. Hence
$f\left(x_{2}-x_{1}\right)>1$.
Thus, $f\left(x_{1}\right) f\left(x_{2}-x_{1}\right)>f\left(x_{1}\right)$, which means
$$
\begin{array}{l}
f\left(x_{2}\right)=f\left(x_{1}+x_{2}-x_{1}\right) \\
=f\left(x_{1}\right) f\left(x_{2}-x_{1}\right)>f\left(x_{1}\right) .
\end{array}
$$

Therefore, $f(x)$ is an increasing function on $\mathbf{R}$.
(4) Solve the inequality $f(x) \leqslant \frac{1}{f(x+1)}$.
$$
\begin{array}{l}
f(x) \leqslant \frac{1}{f(x+1)} \Leftrightarrow f(x) f(x+1) \leqslant 1 \\
\Leftrightarrow f(2 x+1) \leqslant f(0) \Leftrightarrow 2 x+1 \leqslant 0,
\end{array}
$$

Thus, the solution set is $\left\{x \mid x \in \mathbf{R}, x \leqslant-\frac{1}{2}\right\}$.
Note: The key to solving this problem is to identify the prototype of the abstract function, thereby deducing the properties of the abstract function to achieve the purpose of solving the problem.