Example 1 Solve the inequality $\lg \left(x-\frac{1}{x}\right) \leqslant 0$.
Solve The original inequality is equivalent to $0 \leqslant \frac{x^{2}-1}{x} \leqslant 1 \Leftrightarrow \frac{x^{2}-1}{x}\left(\frac{x^{2}-1}{x}-1\right) \leqslant 0 \Leftrightarrow \frac{\left(x^{2}-1\right)\left(x^{2}-1-x\right)}{x^{2}} \leqslant 0 \Leftrightarrow\left\{\begin{array}{l}x \neq 0 \\ (x+1)(x-1)\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right) \leqslant 0,\end{array}\right.$ By the root marking method on the number line (as shown in Figure 2-1), the solution to the given inequality is $-1 \leqslant x \leqslant \frac{1-\sqrt{5}}{2}$ or $1 \leqslant x \leqslant \frac{1+\sqrt{5}}{2}$.
Explanation In the solution process, the following conclusion is used:
$$
a<b \text { when, } a \leqslant f(x) \leqslant b \Leftrightarrow(f(x)-a)(f(x)-b) \leqslant 0 \text {. }
$$

In fact, in the rule of solving quadratic inequalities, there is $(x-a)(x-b) \leqslant 0 \Leftrightarrow a \leqslant x \leqslant b,(a<b)$, so by substituting $f(x)$ for $x$, we have $(f(x)-a)(f(x)-b) \leqslant 0 \Leftrightarrow a \leqslant f(x) \leqslant b$.

Using this conclusion, the double-chain inequality can be converted into a single-direction inequality. In problem solving, if we pay attention to using this conversion relationship, many problems related to double-chain inequalities can be solved surprisingly, thus avoiding the complex operation process of solving systems of inequalities or proving in different directions.