Example 6 (adapted from a problem in *Crux Mathematicorum*) Let $\lambda (\lambda > -1)$ be a real constant, and $x, y, z$ be non-negative real numbers, satisfying $\sum x = 1, \sum xy > 0$. Find the maximum value of $M$ such that the inequality $(1+\lambda x)(1+\lambda y)(1+\lambda z) \geqslant M(1-x)(1-y)(1-z)$ always holds.
Given $x, y, z \in [0,1)$, let $F(x, y, z)=$
$$\frac{(1+\lambda x)(1+\lambda y)(1+\lambda z)}{(1-x)(1-y)(1-z)}=\lambda^{2}+(\lambda+1)$$
$\frac{\lambda^{2} x y z+1}{x y+y z+z x-x y z}$, the problem is to find the minimum value of $F(x, y, z)$.

We conjecture that $F(x, y, z) \geqslant F\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$ holds, i.e.,
$$\left(\lambda^{2}+27\right)(x y+y z+z x-x y z) \leqslant 8\left(\lambda^{2} x y z+1\right)$$
$$\Leftrightarrow \lambda^{2}\left(\sum x y-9 x y z\right)+27\left(\sum x y-x y z\right) \leqslant 8$$

Substituting $\sum x=1$ into $\left(\sum x\right)^{3}+9 x y z \geqslant 4\left(\sum x\right)\left(\sum y z\right)$, we get $4 \sum y z-9 x y z \leqslant 1$,
$$\begin{aligned}
\quad \text { and } \sum y z=\left(\sum y z\right)\left(\sum x\right)=\sum x y(x+y) \\
+3 x y z \geqslant 9 x y z
\end{aligned}$$

When $\lambda^{2} \leqslant 5$, from (2) and (3), we have $8 \geqslant 8\left(4 \sum x y-9 x y z\right) \geqslant \lambda^{2}\left(\sum x y-9 x y z\right)+27\left(\sum x y-x y z\right)$, which means (1) holds. Therefore, when $\lambda^{2} \leqslant 5$, $M_{\text {max }}=F\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)=\frac{(\lambda+3)^{3}}{8}$.

Next, consider $\lambda^{2}>5$, and conjecture that $F(x, y, z) \geqslant F\left(\frac{1}{2}, \frac{1}{2}, 0\right)$ holds, i.e.,
$$4(x y+y z+z x-x y z) \leqslant \lambda^{2} x y z + 1 \Leftrightarrow 4 \sum y z-9 x y z \leqslant\left(\lambda^{2}-5\right) x y z+1$$

When $\lambda^{2}>5$, from (2), we know (4) holds. Therefore, when $\lambda^{2}>5$, $M_{\text {max }}=F\left(\frac{1}{2}, \frac{1}{2}, 0\right)=(\lambda+2)^{2}$.