Example 2 Let $f(n)$ be a positive function defined on the set of natural numbers, $f(1)=1$ and satisfies
$$
f(n-1)-f(n) \geqslant n f(n-1) f(n)(n \geqslant 2) \text {. }
$$

Find $f(n)$.
Analysis: Dividing both sides of equation (1) by $f(n-1) f(n)$ yields
$$
\begin{array}{l}
\frac{1}{f(n)}-\frac{1}{f(n-1)} \geqslant n(n \geqslant 2) . \\
\text { Then } \frac{1}{f(n)}-[n+(n-1)+\cdots+1] \\
\geqslant \frac{1}{f(n-1)}-[(n-1)+\cdots+1] . \\
\text { Let } g(n)=\frac{1}{f(n)}-[n+(n-1)+\cdots+1] .
\end{array}
$$

Then $g(1)=\frac{1}{f(1)}-1=0$.
Thus, the analytical form of the given recursive inequality is
$$
f(n)=\frac{2}{n(n+1)+2 g(n)},
$$

where any function $g(n)$ satisfies
$$
g(1)=0, g(n) \geqslant g(n-1) \text {. }
$$

Note: Generally, for inequalities of the form
$$
\alpha f(n+1) f(n)+\beta f(n+1)+\gamma f(n) \geqslant 0
$$

if for any positive integer $n$, we have
$$
f(n+1) f(n)>0(f(n+1) f(n)<0) \text {, }
$$

it is generally possible to solve by taking the reciprocal, converting it to a first-order linear inequality.