Notice that $1+b+c \approx a^{2}+b+c$, we can consider multiplying both the numerator and the denominator by $1+b+c$, and then perform the bounding
$$\sum \frac{1+b+c}{\left(a^{2}+b+c\right)(1+b+c)} \leqslant \sum \frac{1+b+c}{(a+b+c)^{2}}=\frac{3+2 \sum a}{\left(\sum a\right)^{2}} \leqslant 1$$

Proved.