8. In $\triangle A B C$,
$$
\frac{\sin \frac{A}{2} \cdot \sin \frac{B}{2}+\sin \frac{B}{2} \cdot \sin \frac{C}{2}+\sin \frac{C}{2} \cdot \sin \frac{A}{2}}{\sin A+\sin B+\sin C}
$$

the maximum value is
8. $\frac{\sqrt{3}}{6}$.

In $\triangle ABC$, there is the identity
$$
\sum \tan \frac{A}{2} \cdot \tan \frac{B}{2}=1,
$$

where “$\sum$” denotes the cyclic sum.
$$
\begin{array}{l}
\text { Then } \sum \sin \frac{A}{2} \cdot \sin \frac{B}{2} \\
=\sum \sqrt{\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \cos \frac{A}{2} \cdot \cos \frac{B}{2}} \sqrt{\tan \frac{A}{2} \cdot \tan \frac{B}{2}} \\
\leqslant \sqrt{\left(\sum \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \cos \frac{A}{2} \cdot \cos \frac{B}{2}\right) \sum\left(\tan \frac{A}{2} \cdot \tan \frac{B}{2}\right)} \\
=\frac{1}{2} \sqrt{\sum(\sin A \cdot \sin B)} \\
\leqslant \frac{1}{2} \sqrt{\frac{\left(\sum \sin A\right)^{2}}{3}}=\frac{\sqrt{3}}{6} \sum \sin A . \\
\text { Therefore, } \frac{\sum \sin \frac{A}{2} \cdot \sin \frac{B}{2}}{\sum \sin A} \leqslant \frac{\sqrt{3}}{6} .
\end{array}
$$

The equality holds if and only if $\angle A=\angle B=\angle C=\frac{\pi}{3}$.