Example 3 Solve the inequality
$$
\sqrt{x^{2}-a} \geqslant|x-1|-1 \text {. }
$$
$(2008$, Zhejiang Province High School Mathematics Competition)
Solution: Obviously, when $a=0$, the solution set of the inequality is $x \in \mathbf{R}$.
Next, assume $a \neq 0$. Then
$$
\begin{array}{l}
\sqrt{x^{2}-a} \geqslant|x-1|-1 \\
\Leftrightarrow \sqrt{x^{2}-a}+1 \geqslant|x-1| \\
\Leftrightarrow\left(\sqrt{x^{2}-a}+1\right)^{2} \geqslant(x-1)^{2} \\
\Leftrightarrow 2\left(x+\sqrt{x^{2}-a}\right) \geqslant a \\
\Leftrightarrow \frac{2 a}{x-\sqrt{x^{2}-a}} \geqslant a \\
\Leftrightarrow a\left(\frac{2}{x-\sqrt{x^{2}-a}}-1\right) \geqslant 0 .
\end{array}
$$

When $a>0$,
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow \frac{2}{x-\sqrt{x^{2}-a}} \geqslant 1 \\
\Leftrightarrow 2 \geqslant x-\sqrt{x^{2}-a}, \text { and } x \geqslant \sqrt{a} \\
\Leftrightarrow 2+\sqrt{x^{2}-a} \geqslant x, \text { and } x \geqslant \sqrt{a} \\
\Leftrightarrow\left(2+\sqrt{x^{2}-a}\right)^{2} \geqslant x^{2}, \text { and } x \geqslant \sqrt{a} \\
\Leftrightarrow \sqrt{x^{2}-a} \geqslant \frac{a}{4}-1, \text { and } x \geqslant \sqrt{a} .
\end{array}
$$

If $a \leqslant 4$, then Equation (2) holds.
Thus $x \in[\sqrt{a},+\infty)$.
If $a>4$, then
$$
\begin{array}{l}
\text { Equation (2) } \Leftrightarrow x^{2}-a \geqslant\left(\frac{a}{4}-1\right)^{2}, \text { and } x \geqslant \sqrt{a} \\
\Leftrightarrow x^{2} \geqslant\left(\frac{a}{4}+1\right)^{2}, \text { and } x \geqslant \sqrt{a} \\
\Leftrightarrow x \geqslant \frac{a}{4}+1 .
\end{array}
$$

In summary, when $a \leqslant 0$, $x \in \mathbf{R}$;
when $0<a \leqslant 4$, $x \in[\sqrt{a},+\infty)$;
when $a>4$, $x \in\left[\frac{a}{4}+1,+\infty\right)$.