Example 5 Let $f:[0,+\infty) \rightarrow(-\infty,+\infty)$. And it satisfies:
(1) $f(2)=0$;
(2) For $0 \leqslant x < 2$, the function inequality $f[x f(y)] f(y) \leqslant f(x+y)$ holds. When $x \in [0,2)$, try to find $f(x)$.

保留了原文的换行和格式，如上所示。
(1) When $x>2$, let $x=t+2(t>0)$, then
$$
f[t f(2)] f(2) \leqslant f(t+2)=f(x)
$$

Given $f(2)=0$, we know $f(x) \geqslant 0(x \geqslant 2)$.
(2) When $0 \leqslant x0)$. Then
$$
0=f(2)=f(x+t) \geqslant f[t f(x)] f(x)
$$

Since when $0 \leqslant x0, f(x) \leqslant 0$, it follows that $f(x) \geqslant \frac{2}{2-x}$ does not hold, thus the solution to the functional inequality is $f(x)=0, x \in[0,2)$.