Example 2 Given the functional inequality
$$
f\left(x^{2}\right)-f(x) \geqslant 1,
$$

try to find the continuous function $f(x)$, where the domain of $f(x)$ is $(1,+\infty)$.
Construct the sequence $\left\{x_{n}\right\}$ as follows:
$$
x_{0}, x_{1}=x_{11}^{2}, x_{2}=x_{1}^{2}, \cdots, x=x_{n}=x_{n-1}^{2},
$$

Substitute them into the given functional inequality, we get
$$
f\left(x_{1}\right)-f\left(x_{0}\right) \geqslant 1, f\left(x_{2}\right)-f\left(x_{1}\right) \geqslant 1, \cdots, f(x)-f\left(x_{n-1}\right) \geqslant 1,
$$

Adding these inequalities, we get $f(x)-f\left(x_{11}\right) \geqslant n$.
Since $x=x_{0}^{2}$, we have $\log _{2} \log _{2} x-\log _{2} \log _{2} x_{0}=n$,
Therefore, $f(x) \geqslant f\left(x_{0}\right)+\log _{2} \log _{2} x-\log _{2} \log _{2} x_{n}$,
Let $c=f\left(x_{1}\right)-\log _{2} \log _{2} x_{0}$, then $f(x) \geqslant \log _{2} \log _{2} x+c$. It is easy to verify that a particular solution is $f(x)=$ $\log _{2} \log _{2} x+c$.

Analysis Here, the solution $f(x)=\log _{2} \log _{2} x+c$ obtained by iteration is just a particular solution to the functional inequality (1). It is easy to verify that
$$
f(x)=\log _{2} \log _{2} x+\sin \left(2 \pi \log _{2} \log _{2} x\right)
$$

is also a particular solution to the functional inequality (1).
Proof
$$
\begin{aligned}
f\left(x^{2}\right) & =\log _{2} \log _{2} x^{2}+\sin \left(2 \pi \log _{2} \log _{2} x^{2}\right) \\
& =\log _{2}\left(2 \log _{2} x\right)+\sin 2 \pi \log _{2}\left(2 \log _{2} x\right) \\
& =1+\log _{2} \log _{2} x+\sin 2 \pi\left(1+\log _{2} \log _{2} x\right) \\
& =1+\log _{2} \log _{2} x+\sin \left(2 \pi \log _{2} \log _{2} x\right) \\
& =1+f(x) .
\end{aligned}
$$

So $f\left(x^{2}\right)-f(x)=1$, hence $f(x)=\log _{2} \log _{2} x+\sin \left(2 \pi \log _{2} \log _{2} x\right)$ is also a particular solution to the functional inequality (1).
The structure of the two particular solutions suggests that we can try to find a general solution to the functional inequality (1) using the substitution method.

Solution Let $f(x)=g\left(\log _{2} \log _{2} x\right)$, then the functional inequality (1) becomes
$$
g\left(\log _{2} \log _{2} x+1\right)-g\left(\log _{2} \log _{2} x\right) \geqslant 1,
$$

Let $t=\log _{2} \log _{2} x$, since the domain of $f(x)$ is $(1,+\infty)$, we have $t \in \mathbf{R}$. The above equation can be rewritten as
$$
g(t+1)-g(t) \geqslant 1,
$$

which is $g(t+1)-(t+1) \geqslant g(t)-t$. Let $h(t)=g(t)-t$, then $h(t+1) \geqslant h(t)$. By back-substitution, $g(t)=h(t)+t$, the general solution to the original functional inequality (1) is
$$
f(x)=g\left(\log _{2} \log _{2} x\right)=h\left(\log _{2} \log _{2} x\right)+\log _{2} \log _{2} x .
$$

where $h(x)$ is any function defined on $\mathbf{R}$ that satisfies $h(x+1) \geqslant h(x)$.
Note For functional inequalities of the form $f\left(x^{p}\right)-f\left(x^{v}\right) \geqslant 1$, a similar method can be used to find their general solutions.