Example 3 Let $\forall x, y \in \mathbf{R}$, if $f(1)=0, \alpha(y)$ is a constant function, try to find the continuous solutions of the functional inequality
$$
f(x y) \geqslant \alpha(y) f(x)+f(y)
$$
From the given inequality, we have $f(x y)-f(y) \geqslant \alpha(y) f(x)$.
When $x>1$, $\frac{f(x y)-f(y)}{x y} y \geqslant \alpha(y) \frac{f(x)}{x-1}$,
Let $x \rightarrow 1$, we get $f^{\prime}(y) y \geqslant \alpha(y) f^{\prime}(1)$.
When $x<1$, $\frac{f(x y)-f(y)}{x y-y} y \leqslant \alpha(y) \frac{f(x)}{x-1}$,
Let $x \rightarrow 1$, we get $f^{\prime}(y) y \leqslant \alpha(y) f^{\prime}(1)$.
Thus, we have $f^{\prime}(y) y=\alpha(y) f^{\prime}(1)$,
The formal solution to the given functional inequality is $f(y)=f^{\prime}(1) \int_{1}^{y} \frac{\alpha(y)}{y} \mathrm{~d} y$.