Example 1 Given that $f(x)$ is a non-zero function defined on $\mathbf{R}$, and for any $a, b \in \mathbf{R}^{+}$ it satisfies $f(1)=0, f(a \cdot b) \geqslant a f(b)+b f(a)$. Try to determine the relationship between $f\left(a^{n}\right)$ and $f(a)$.
Solution 1:
$$
\begin{array}{c}
f\left(a^{2}\right) \geqslant a f(a)+a f(a)=2 a f(a), \\
f\left(a^{3}\right) \geqslant a^{2} f(a)+a f\left(a^{2}\right) \geqslant 3 a^{2} f(a),
\end{array}
$$

Guess $f\left(a^{n}\right) \geqslant n a^{n-1} f(a)$.
We will prove this by mathematical induction:
1. When $n=1$, $f\left(a^{1}\right) \geqslant 1 \cdot a^{n} \cdot f(a)$, the inequality holds;
2. Assume that when $n=k$, the inequality $f\left(a^{k}\right) \geqslant k a^{k-1} f(a)$ holds,

Then when $n=k+1$,
$$
\begin{aligned}
f\left(a^{k+1}\right) & \geqslant a^{k} f(a)+a f\left(a^{k}\right) \\
& \geqslant a^{k} f(a)+k a^{k} f(a) \\
& =(k+1) a^{k} f(a), \text{ which still holds. }
\end{aligned}
$$

From the above two steps, we can conclude that for any $n \in \mathbf{N}^{+}, f\left(a^{n}\right) \geqslant n a^{n-1} f(a)$ holds.
Solution 2: When $a b \neq 0$, $\frac{f(a \cdot b)}{a b} \geqslant \frac{f(b)}{b}+\frac{f(a)}{a}$,
Let $g(x)=\frac{f(x)}{x}$, then $g(a \cdot b) \geqslant g(a)+g(b)$,
Thus $g\left(a^{n}\right) \geqslant n g(a)$. Therefore, $f\left(a^{n}\right) \geqslant a^{n} \cdot g\left(a^{n}\right)=n a^{n} g(a)=n a^{n-1} f(a)$.