Example 9 For positive real numbers $a, b, c$ satisfying $abc=1$, find the maximum value of
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right)$$
Given that the expression is symmetric with respect to $a$, $b$, and $c$, when $a=b=c=1$, we have
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right)=1$$

Next, we prove that the maximum value is 1, i.e., for any real numbers $a$, $b$, and $c$ satisfying $a b c=1$, we have
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$$

First, we convert the non-homogeneous expression into a homogeneous one. For positive real numbers $x, y, z$, let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ (for example, $x=1, y=\frac{1}{a}, z=\frac{1}{a b}$). Then the above inequality is equivalent to proving
$$(x-y+z)(y-z+x)(z-x+y) \leqslant x y z$$

Let $u=x-y+z, v=y-z+x, w=z-x+y$. Since the sum of any two of $u, v, w$ is positive, at most one of them can be negative. Therefore, without loss of generality, assume $u \geqslant 0, v \geqslant 0$. By the AM-GM inequality, we have
$$\begin{aligned}
\sqrt{u v} & =\sqrt{(x-y+z)(y-z+x)} \\
& \leqslant \frac{1}{2}(x-y+z+y-z+x)=x
\end{aligned}$$

Similarly, $\sqrt{v w} \leqslant y, \sqrt{w u} \leqslant z$, hence $u v w \leqslant x y z$.