$$\begin{aligned} & \sum_{\text {cyc }} \frac{1}{1+a^{2}(b+c)} \leqslant \frac{3}{1+2 a b c} \Leftrightarrow \sum_{\text {cyc }}\left(\frac{1}{1+2 a b c}-\frac{1}{1+a^{2}(b+c)}\right) \geqslant 0 \\ \Leftrightarrow & \sum_{\text {cyc }} \frac{a^{2}(b+c)-2 a b c}{1+a^{2}(b+c)} \geqslant 0 \Leftrightarrow \sum_{\text {cyc }} \frac{a c(a-b)-a b(c-a)}{1+a^{2}(b+c)} \geqslant 0 \\ \Leftrightarrow & \sum_{\text {cyc }}(a-b)\left(\frac{a c}{1+a^{2}(b+c)}-\frac{b c}{1+b^{2}(a+c)}\right) \geqslant 0 \\ \Leftrightarrow & \sum_{\text {cyc }} \frac{c(1-a b c)(a-b)^{2}}{\left(1+a^{2}(b+c)\right)\left(1+b^{2}(a+c)\right)} \geqslant 0 . \end{aligned}$$

In fact, by $\frac{b c+c a+a b}{3} \geqslant \sqrt[3]{a^{2} b^{2} c^{2}}$, we know that $1-a b c \geqslant 0$.
Therefore, equation (8) holds.