Example 1 Let $f$ be a positive-valued function defined on the set of positive integers, with $f(1)=1$ and
$$
f(n+1) \geqslant(n+1) f(n)-2 n \quad (n \geqslant 1) .
$$

Try to find $f(n)$.
Analysis: Dividing both sides of equation (1) by $(n+1)!$ yields
$$
\begin{array}{l}
\frac{f(n+1)}{(n+1)!} \geqslant \frac{f(n)}{n!}-\frac{2 n}{(n+1)!} \\
=\frac{f(n)}{n!}-\frac{2(n+1-1)}{(n+1)!} \\
=\frac{f(n)}{n!}-2\left[\frac{1}{n!}-\frac{1}{(n+1)!}\right],
\end{array}
$$

which means $\frac{f(n+1)-2}{(n+1)!} \geqslant \frac{f(n)-2}{n!}$.
Let $\frac{f(n)-2}{n!}=g(n)$. Then
$$
f(n)=n!g(n)-2 \text {, }
$$
$$
\begin{array}{l}
g(1)=\frac{f(1)-2}{1!}=-1, \\
g(n+1) \geqslant g(n) .
\end{array}
$$
Solution: Let $f(n)=n!g(n)-2$. Substituting into equation (1) yields
$$
\begin{array}{l}
(n+1)!g(n+1)-2 \\
\geqslant(n+1)(n!g(n)-2)-2 n .
\end{array}
$$

Simplifying, we get $g(n+1) \geqslant g(n)$.
Thus, the analytical expression for the given recursive inequality is
$$
f(n)=n!g(n)-2 \text {, }
$$

where any function $g(n)$ satisfies
$$
g(1)=\frac{f(1)-2}{1!}=-1, g(n+1) \geqslant g(n) \text {. }
$$

Note: Generally, let $f(n)$ be a non-zero function. Then the general term formula for the sequence $\{f(n)\}$ that satisfies the first-order linear recursive inequality
$$
f(n+1) \geqslant a_{n} f(n)+d(n)
$$

can be found using the "successive summation method":

Divide both sides of equation (1) by $a_{n} a_{n-1} \cdots a_{2} a_{1}$ (or its absolute value $\left|a_{n} a_{n-1} \cdots a_{2} a_{1}\right|$, yielding
$$
\begin{array}{l}
\frac{f(n+1)}{a_{n} a_{n-1} \cdots a_{2} a_{1}} \geqslant \frac{f(n)}{a_{n-1} \cdots a_{2} a_{1}}+\frac{d(n)}{a_{n} a_{n-1} \cdots a_{2} a_{1}} . \\
\text { Let } b_{n}=\frac{f(n)}{a_{n-1} \cdots a_{2} a_{1}} \text {. Then } \\
b_{n+1} \geqslant b_{n}+\frac{d(n)}{a_{n} a_{n-1} \cdots a_{2} a_{1}} .
\end{array}
$$