Prove
$$\begin{array}{l} 4 \sum x \cdot \sum \frac{y z}{x(y+z)^{2}}-9=\left(4 \sum \frac{y z}{(y+z)^{2}}-3\right)+\left(4 \sum \frac{y z}{x(y+z)}-6\right)= \\ 2 \sum \frac{x^{2}(y-z)^{2}}{y z(x+y)(x+z)}-\sum\left(\frac{y-z}{y+z}\right)^{2}= \\ \frac{1}{\prod(y+z)} \cdot \sum \frac{3 x^{2} y z+2 x^{2} y^{2}+2 x^{2} z^{2}-x y^{2} z-x y z^{2}-y^{2} z^{2}}{y z(y+z)}(y-z)^{2} \end{array}$$

By symmetry, without loss of generality, assume $x \geqslant y \geqslant z$. It is easy to prove that
$$\frac{(x-z)^{2}}{x z(x+z)} \geqslant \frac{(x-y)^{2}}{x y(x+y)}$$

And
$$\begin{array}{l} 3 x^{2} y z+2 x^{2} y^{2}+2 x^{2} z^{2}-x y^{2} z-x y z^{2}-y^{2} z^{2}>0 \\ 3 x y^{2} z+2 x^{2} y^{2}+2 y^{2} z^{2}-x^{2} y z-x y z^{2}-x^{2} z^{2}>0 \end{array}$$

Thus, we have
$$\begin{array}{l} \sum \frac{3 x^{2} y z+2 x^{2} y^{2}+2 x^{2} z^{2}-x y^{2} z-x y z^{2}-y^{2} z^{2}}{y z(y+z)}(y-z)^{2} \geqslant \\ {\left[\left(3 x y^{2} z+2 x^{2} y^{2}+2 y^{2} z^{2}-x^{2} y z-x y z^{2}-x^{2} z^{2}\right)+\right.} \\ \left.\left(3 x y z^{2}+2 x^{2} z^{2}+2 y^{2} z^{2}-x^{2} y z-x y^{2} z-x^{2} y^{2}\right)\right] \cdot \frac{(x-y)^{2}}{x y(x+y)}= \\ \left(2 x y^{2} z+2 x y z^{2}-2 x^{2} y z+x^{2} y^{2}+x^{2} z^{2}+4 y^{2} z^{2}\right) \cdot \frac{(x-y)^{2}}{x y(x+y)} \geqslant \end{array}$$
0
Note: Equation (4) is equivalent to $a, b, c \in \mathbf{R}^{+}$, then $\sum b c \cdot \sum \frac{1}{(b+c)^{2}} \geqslant \frac{9}{4}$. Refer to Example 19 in Chapter 7 "Other Inequality Proof Examples".