Example 4 (Crux2006: $415-416$ ) Let $k>-1$ be a fixed real number, and $a, b, c$ be non-negative real numbers, satisfying $a+b+c=1$, and $ab + bc + ca > 0$. Find: $\min \left\{\frac{(1+ka)(1+kb)(1+kc)}{(1-a)(1-b)(1-c)}\right\}$.
Solution: The required minimum value is $\min \left\{\frac{1}{8}(k+3)^{3},(k+2)^{2}\right\}$. First, we prove the following inequality:
$4(a b+b c+c a) \leqslant 1+9 a b c$.
Using the condition $a+b+c=1$, we get
$$\begin{array}{l}
1-4(a b+b c+c a)+9 a b c \\
=(a+b+c)^{3}-4(a+b+c)(a b+b c+c a)+9 a b c \\
=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c \\
\quad-b) \geqslant 0
\end{array}$$

The latter part of the above expression is Schur's inequality (2), which proves that (4) holds.

We also assert that $a b+b c+c a \geqslant 9 a b c$.
In fact, using the AM-GM inequality, we get
$$\begin{array}{l}
a b+b c+c a=(a b+b c+c a)(a+b+c) \\
=a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}+3 a b c \\
\geqslant 6 a b c+3 a b c=9 a b c
\end{array}$$

Returning to the original problem, we find that $a, b$, or $c$ cannot be equal to 1. For example, if $a=1$, then $b=c=0$, which means $a b+b c+c a=0$, a contradiction. Thus, $a, b, c \in[0,1)$, and let
$$\begin{array}{l}
Q(a, b, c)=\frac{(1+k a)(1+k b)(1+k c)}{(1-a)(1-b)(1-c)} \\
=\frac{k^{3} a b c+k^{2}(a b+b c+c a)+k+1}{a b+b c+c a-a b c} \\
=k^{2}+(k+1) \cdot \frac{k^{2} a b c+1}{a b+b c+c a-a b c}
\end{array}$$

Notice that $Q\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)=\frac{1}{8}(k+3)^{3}$ and $Q\left(0, \frac{1}{2}, \frac{1}{2}\right)$ $=(k+2)^{2}$.

Case 1: When $k^{2} \leqslant 5$.
We prove that $Q(a, b, c) \geqslant Q\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.
Since $k+1>0$, direct calculation yields
$$k^{2}(a b+b c+c a-9 a b c)+27(a b+b c+c a-a b c)$$
$\leqslant 8$.
By (5), we know that $k^{2}(a b+b c+c a-9 a b c) \geqslant 0$. Since $k^{2} \leqslant 5$, the left side of (6) is at most $8[4(a b+b c+c a)-9 a b c]$, so by (4), (6) holds.

Thus, $Q\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)=\frac{1}{8}(k+3)^{3}$ is the minimum value of $Q(a, b, c)$.

Case 2: When $k^{2} \geqslant 5$.
We prove that $Q(a, b, c) \geqslant Q\left(0, \frac{1}{2}, \frac{1}{2}\right)$.
Since $k+1>0$, we deduce that this inequality is equivalent to
$1+4[a b c-(a b+b c+c a)]+k^{2} a b c \geqslant 0$.
Since $k^{2} \geqslant 5$, by (4), this inequality holds. Thus, $Q\left(0, \frac{1}{2}, \frac{1}{2}\right)=(k+2)^{2}$ is the minimum value of $Q(a, b, c)$.

Notice that $\frac{1}{8}(k+3)^{3}-(k+2)^{2}=\frac{1}{8}(k+1)\left(k^{2}\right.$ $-5)$, we find that if $k^{2} \geqslant 5$, then $\frac{1}{8}(k+3)^{3} \geqslant(k+2)^{2}$; if $k^{2} \leqslant 5$, then $\frac{1}{8}(k+3)^{3} \leqslant(k+2)^{2}$. The conclusion holds.