67th Romanian Mathematical Olympiad

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 $S\ge 9-2(a+b+c+1)S\ge 1$.