Analysis 1
$$b-1+\frac{1}{c}=b\left(1-\frac{1}{b}+\frac{1}{b c}\right)=b\left(1+a-\frac{1}{b}\right),$$

Therefore,
$$\begin{aligned} \left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right) & =b\left(a-1+\frac{1}{b}\right)\left(a+1-\frac{1}{b}\right) \\ & =b\left[a^{2}-\left(1-\frac{1}{b}\right)^{2}\right] \leqslant b a^{2}, \end{aligned}$$

Similarly, $\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant c b^{2},\left(c-1+\frac{1}{a}\right)\left(a-1+\frac{1}{b}\right) \leqslant a c^{2}$. If $a-1+\frac{1}{b}, b-1+\frac{1}{c}, c-1+\frac{1}{a}$ are not all positive, without loss of generality, assume $a-1+\frac{1}{b} \leqslant 0$, then $a \leqslant 1-\frac{1}{b}0, c-1+\frac{1}{a}>0$, hence the proposition holds.
If $a-1+\frac{1}{b}, b-1+\frac{1}{c}, c-1+\frac{1}{a}$ are all positive, then from the above three inequalities we get
$$\begin{array}{l} \left(a-1+\frac{1}{b}\right)^{2}\left(b-1+\frac{1}{c}\right)^{2}\left(c-1+\frac{1}{a}\right)^{2} \leqslant a^{3} b^{3} c^{3}=1, \\ \left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 . \end{array}$$

Analysis 2
By $a b c=1$, without loss of generality, let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, where $x, y, z \in \mathbf{R}^{+}$. Then the inequality to be proved is
$$\left(\frac{x}{y}-1+\frac{z}{y}\right)\left(\frac{y}{z}-1+\frac{x}{z}\right)\left(\frac{z}{x}-1+\frac{y}{x}\right) \leqslant 1,$$

which is
$$(x+y-z)(y+z-x)(z+x-y) \leqslant x y z .$$

Since the sum of any two of $x+y-z, y+z-x, z+x-y$ is positive, at most one of them can be negative. If exactly one is negative, then (1) holds; if all three are non-negative, then
$$\begin{array}{l} x^{2} \geqslant x^{2}-(y-z)^{2}=(x+y-z)(x+z-y), \\ y^{2} \geqslant(y+z-x)(x+y-z), z^{2} \geqslant(z+x-y)(y+z-x), \end{array}$$

Multiplying these three inequalities and taking the square root, we get $(x+y-z)(y+z-x)(z+x-y) \leqslant x y z$. Hence the proposition is proved.