Example 5.17 Given non-negative real numbers $a, b, c$ satisfy $a b + b c + c a + 6 a b c = 9$. Determine the maximum value of $k$ such that the following inequality always holds.
$$a + b + c + k a b c \geqslant k + 3$$
Let $a=b=3, c=0$, then we have $k \leqslant 3$. Now, let's prove that
$$a+b+c+3 a b c \geqslant 6$$

Let $p=a+b+c, q=a b+b c+c a, r=a b c$, then from the problem, we have
$$9=q+6 r \geqslant 3 r^{\frac{2}{3}}+6 r \Rightarrow r \leqslant 1 \Rightarrow q=9-6 r \geqslant 3$$

And $p^{2} \geqslant 3 q$, so $p \geqslant 3$. To prove $p+3 r \geqslant 6$, it is equivalent to proving $2 p-q \geqslant 3$. We will discuss this in cases.
(1) If $p \geqslant 6$, then the original inequality obviously holds.
(2) If $3 \leqslant p \leqslant 6$.
(1) If $p^{2} \geqslant 4 q$, then
$$2 p-q \geqslant 2 p-\frac{p^{2}}{4}=\frac{(p-2)(6-p)}{4}+3 \geqslant 3$$
(2) If $p^{2} \leqslant 4 q$, since $r \geqslant \frac{p\left(4 q-p^{2}\right)}{9}$, then
$$\begin{array}{c}
27=3 q+18 r \geqslant 3 q+2 p\left(4 q-p^{2}\right) \Leftrightarrow \\
(p+1)(p-3)(p-6) \leqslant 0
\end{array}$$

Thus, we have $k_{\max }=3$.