Prove that
$$\begin{array}{l} (x y+y z+z x)\left(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right)= \\ (x y+y z+x z) \frac{(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}}{(x+y)^{2}(y+z)^{2}(z+x)^{2}} \end{array}$$

But we have
$$\begin{array}{l} (x y+y z+z x)\left((x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}\right)= \\ \sum\left(x^{5} y+2 x^{4} y^{2}+\frac{5}{2} x^{4} y z+13 x^{3} y^{2} z+4 x^{2} y^{2} z^{2}\right) \\ (x+y)^{2}(y+z)^{2}(z+x)^{2}=\sum_{y m}\left(x^{4} y^{2}+x^{4} y z+x^{3} y^{3}+6 x^{3} y^{2} z+\frac{5}{3} x^{2} y^{2} z^{2}\right) \end{array}$$

Through some calculations, we have
$$\sum_{s y m}\left(4 x^{5} y-x^{4} y^{2}-3 x^{3} y^{3}+x^{4} y z-2 x^{3} y^{2} z+x^{2} y^{2} z^{2}\right) \geqslant 0$$

By the 3rd degree Schur's inequality, we have
$$\sum_{s y m}\left(x^{3}-2 x^{2} y+x y z\right) \geqslant 0$$

Multiplying both sides by $x y z$, we get
$$\sum_{s y m}\left(x^{4} y z-2 x^{3} y^{2} z+x^{2} y^{2} z^{2}\right) \geqslant 0$$

By the AM-GM inequality, we have
$$\sum_{y>m}\left(\left(x^{5} y-x^{4} y^{2}\right)+3\left(x^{5} y-x^{3} y^{3}\right)\right) \geqslant 0$$

Using (1) and (2), the proposition is proved.