Proof
$$\Leftrightarrow \sum \frac{1}{4-(b+c)^{2}} \leqslant \frac{9}{8} \Leftrightarrow \sum \frac{(b+c)^{2}}{4-(b+c)^{2}} \leqslant \frac{3}{2}$$

By Cauchy inequality, we have
$$\begin{array}{l} \sum \frac{(b+c)^{2}}{4-(b+c)^{2}} \leqslant \sum \frac{(b+c)^{2}}{4-2\left(b^{2}+c^{2}\right)}= \\ \frac{1}{2} \sum \frac{(b+c)^{2}}{\left(a^{2}+b^{2}\right)+\left(a^{2}+c^{2}\right)} \leqslant \\ \frac{1}{2} \sum\left(\frac{b^{2}}{a^{2}+b^{2}}+\frac{c^{2}}{a^{2}+c^{2}}\right)=\frac{3}{2} \end{array}$$

Hence, it is proved!