Example 1 Given $f(\sqrt{x}+1) \geqslant x+2 \sqrt{x}$, try to find $f(x)$.

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Solution: Since $f(\sqrt{x}+1) \geqslant(\sqrt{x}+1)^{2}-1$. Therefore, $f(x) \geqslant x^{2}-1 \quad(x \geqslant 1)$.
Note: In fact, we have only found the solution to the functional inequality under the constraint. For $x<1$, the given functional inequality does not impose any constraint on $f(x)$, so it can take any value, or it can be undefined.
The following functions are all solutions to the given functional inequality:
(1) $f(x) \geqslant x^{2}-1$;
(2) $f(x) \geqslant x^{2}-1, x \in[0,+\infty)$;
(3) $f(x) \geqslant x^{2}-1, x \in\{0\} \cup[1,+\infty)$;
(4) $f(x)=\left\{\begin{array}{l}x^{2}-1, x \geqslant 1 \\ x, x<1\end{array}\right.$;
(5) $f(x)=\left\{\begin{array}{l}x^{2}-1, x \geqslant 1 \\ \sin x, x<1\end{array}\right.$;
(6) $f(x)=\left\{\begin{array}{l}x^{2}-1, x \geqslant 1 \\ 2^{r},-10<x<-1\end{array}\right.$.