Prove that due to the symmetry of the inequality about $a, b, c$, without loss of generality, assume $0 \leqslant a \leqslant b \leqslant c \leqslant 1$, then
$$\begin{aligned} \text { LHS }= & \sum \frac{a}{b+c+1}+(1-a)(1-b)(1-c) \leqslant \\ & \frac{a+b+c}{a+b+1}+(1-a)(1-b)(1-c)= \\ & 1-\frac{1-c}{a+b+1}[1-(1+a+b)(1-a)(1-b)] \end{aligned}$$

It suffices to prove
$$\frac{1-c}{a+b+1}[1-(1+a+b)(1-a)(1-b)] \geqslant 0$$

In fact
$$\begin{aligned} (1+a+b)(1-a)(1-b) \leqslant & (1+a+b+ab)(1-a)(1-b)= \\ & \left(1-a^{2}\right)\left(1-b^{2}\right) \leqslant 1 \end{aligned}$$

Thus, the original inequality holds.
Equality holds if and only if $a=b=c=1$ or $a=1, b=c=0$ and their cyclic permutations.