Example 1 Given the function $f(x)=\log _{2}\left(x^{2}+1\right)(x \geqslant 0), g(x)=\sqrt{x-a}(a \in \mathbf{R})$.
(1) Try to find the inverse function $f^{-1}(x)$ of the function $f(x)$;
(2) The function $h(x)=f^{-1}(x)+g(x)$, find the domain of the function $h(x)$, and determine the monotonicity of $h(x)$;
(3) If the function $h(x)$ in (2) satisfies $h(x) \geqslant 2$ for all $x$ in its domain, find the range of the real number $a$.
(1) Since $x \geqslant 0$, we have $f(x)=\log _{2}\left(x^{2}+1\right) \geqslant 0$.
From $y=f(x)=\log _{2}\left(x^{2}+1\right)$, we get $x^{2}=2^{y}-1$, i.e., $x=\sqrt{2^{y}-1}$. Therefore,
$$
f^{-1}(x)=\sqrt{2^{x}-1}(x \geqslant 0) \text {. }
$$
(2) $h(x)=\sqrt{2^{x}-1}+\sqrt{x-a}(x \geqslant 0$ and $x \geqslant a)$.

When $a<0$, the domain is $[0,+\infty)$; when $a \geqslant 0$, the domain is $[a,+\infty)$.
$h(x)$ is monotonically increasing in its domain.
(3) When $a<0$, the domain is $[0,+\infty)$, and $h(x)$ is monotonically increasing in its domain. Then $\left\{\begin{array}{l}a<0, \\ h(0) \geqslant 2,\end{array}\right.$ which gives $a \leqslant-4$.

When $a \geqslant 0$, the domain is $[a,+\infty)$, and $h(x)$ is monotonically increasing in its domain. Then $\left\{\begin{array}{l}a \geqslant 0, \\ h(a) \geqslant 2,\end{array}\right.$ which gives $a \geqslant \log _{2} 5$.
In summary, we have $(-\infty,-4] \cup\left[\log _{2} 5,+\infty\right)$.