Denote by $S = \sum \frac{1}{b+c+1}$. Using the inequality from the hypothesis we obtain
$$\sum \left( \frac{a}{b+c+1} + 1 \right) \le 4,$$
and thus $(a+b+c+1)S \le 4$.

Using arithmetic mean – harmonic mean inequality we get
$$S \ge \frac{9}{2(a + b + c + 1) + 1},$$
hence $S \ge 9 - 2(a + b + c + 1)S \ge 1$.