Example 3 Solve the inequalities: (1) $\sqrt{x-2}<x-2$; (2) $\sqrt{x-1} \geqslant x-3$.
$$
\begin{array}{l}
\text { (1) Since } \sqrt{x-2}>0, \\
12, \\
1<x-2 .
\end{array}\right.
\end{array}
$$

Therefore, the solution set of the original inequality is $(3,+\infty)$.
(2) Let $t=\sqrt{x-1}$, then $t \geqslant 0, x=t^{2}+1$.

The original inequality becomes $t \geqslant t^{2}-2$, solving this yields $-1 \leqslant t \leqslant 2$, that is
$$
\sqrt{x-1} \leqslant 2 \Rightarrow\left\{\begin{array}{l}
x-1 \geqslant 0, \\
x-1 \leqslant 4 .
\end{array}\right.
$$

Therefore, the solution set of the original inequality is $[1,5]$.
Key points for solving the problem: The two methods used in this problem have certain particularities, both related to the characteristics of the inequalities being solved.
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