Let $\frac{A I}{A A^{\prime}}=x, \frac{B I}{B B^{\prime}}=y, \frac{C I}{C C^{\prime}}=z$. By the Angle Bisector Theorem, we have
$$\begin{array}{l} A^{\prime} C=\frac{a b}{b+c}, \\ x=\frac{b}{b+A^{\prime} C}=\frac{b+c}{a+b+c} \end{array}$$

Similarly, $y=\frac{c+a}{a+b+c}$,
$$z=\frac{a+b}{a+b+c}$$

Thus, $x+y+z=2$. Therefore,
$$x y z \leqslant\left[\frac{1}{3}(x+y+z)\right]^{3} \leqslant \frac{8}{27},$$

with equality if and only if $\triangle A B C$ is an equilateral triangle.
$$\text { Also, } x=\frac{2(b+c)}{2(a+b+c)}>\frac{b+c+a}{2(a+b+c)}=\frac{1}{2} \text {, }$$

Similarly, $y>\frac{1}{2}, z>\frac{1}{2}$.
Let $x=\frac{1+\varepsilon_{1}}{2}, y=\frac{1+\varepsilon_{2}}{2}, z=\frac{1+\varepsilon_{3}}{2}, \varepsilon_{1}, \varepsilon_{2}$, and $\varepsilon_{3}$ are all positive, and $\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}=1$,
$$\begin{array}{l} x y z=\frac{\left(1+\varepsilon_{1}\right)\left(1+\varepsilon_{2}\right)\left(1+\varepsilon_{3}\right)}{8} \\ >\frac{1+\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}}{8}=\frac{1}{4} . \end{array}$$

The proposition is proved.