Example 4 Let $f(1)=1, f(2)=2$ and satisfy:
$$
f(n+2) \geqslant 2 f(n+1)-f(n), \forall n \in \mathbf{N}^{+} .
$$

Try to find $f(n)$.
From (1), we know that
$$
f(n+2)-f(n+1) \geqslant f(n+1)-f(n).
$$

Given $f(1)=1 \cdot f(2)=2$, it follows that $f(n+1)-f(n)>0, n \in \mathbf{N}^{+}$.
Thus, we can set
$$
f(n+2)-f(n+1)=g(n)[f(n+1)-f(n)], \forall g(n) \geqslant 1
$$

Substituting $n=1,2,3, \cdots, n-2$ into the above equation, we get
$$
\begin{array}{l}
f(3)-f(2)=g(1)[f(2)-f(1)], \\
f(4)-f(3)=g(2)[f(3)-f(2)], \\
\cdots \cdots \\
f(n)-f(n-1)=g(n-2)[f(n-1)-f(n-2)],
\end{array}
$$

Multiplying these $n-2$ equations, we obtain
$$
\begin{aligned}
f(n)-f(n-1) & =g(1) g(2) \cdots g(n-2)[f(2)-f(1)] \\
& =g(1) g(2) \cdots g(n-2)(2-1), \\
f(n)-f(n-1) & =g(1) g(2) \cdots g(n-2).
\end{aligned}
$$

Substituting $n=2,3,4, \cdots, n$ into (2), we get
$$
\begin{array}{l}
f(2)-f(1)=1, \\
f(3)-f(2)=g(1), \\
f(4)-f(3)=g(1) g(2), \\
\cdots \cdots \\
f(n)-f(n-1)=g(1) g(2) \cdots g(n-2),
\end{array}
$$

Adding these $n-1$ equations, we get
$$
\begin{array}{c}
f(n)-f(1)=1+g(1)+g(1) g(2)+\cdots+g(1) g(2) \cdots g(n-2), \\
f(n)=2+g(1)+g(1) g(2)+\cdots+g(1) g(2) \cdots g(n-2), \forall g(n) \geqslant 1, n \in \mathbf{N}^{+}.
\end{array}
$$