Four. (20 points) Let $x, y, z$ be the lengths of the sides of a triangle, and $x+y+z=1$. Find the minimum value of the real number $\lambda$ such that
$$
\lambda(x y+y z+z x) \geqslant 3(\lambda+1) x y z+1
$$

always holds.
From the original inequality, we have
$$
\lambda \geqslant \frac{3 x y z+1}{x y+y z+z x-3 x y z} \text {. }
$$

When $x=y=z=\frac{1}{3}$, $\lambda \geqslant 5$.
Next, we prove that the minimum value of $\lambda$ is 5, i.e., we need to prove
$$
5(x y+y z+z x) \geqslant 18 x y z+1 \text {. }
$$

Let $x=a+b, y=b+c, z=c+a$.
Then $a+b+c=\frac{1}{2}$, and $a, b, c>0$.
Thus, $5(x y+y z+z x)$
$$
\begin{array}{l}
=5[(a+b)(b+c)+(b+c)(c+a)+ \\
(c+a)(a+b)] \\
=5\left[\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-a\right)+\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)+\right. \\
\left.\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right)\right] \\
\geqslant 18\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)+1 \\
\Leftrightarrow 5\left(\frac{1}{4}+a b+b c+c a\right) \\
\geqslant 9(a b+b c+c a)-18 a b c+1 \\
\Leftrightarrow \frac{1}{4} \geqslant 4(a b+b c+c a)-18 a b c \text {. } \\
\end{array}
$$

For any positive numbers $a, b, c$, we have
$$
(a+b-c)(b+c-a)(c+a-b) \leqslant a b c \text {, }
$$

and since $a+b+c=\frac{1}{2}$, therefore,
$$
\begin{array}{l}
\left(\frac{1}{2}-2 c\right)\left(\frac{1}{2}-2 a\right)\left(\frac{1}{2}-2 b\right) \leqslant a b c \\
\Rightarrow 2(a b+b c+c a) \leqslant 9 a b c+\frac{1}{8} \\
\Rightarrow \frac{1}{4} \geqslant 4(a b+b c+c a)-18 a b c
\end{array}
$$