First of all, if $x$, $y > -1$, then
$$\frac{x}{1+x} < \frac{y}{1+y} \Leftrightarrow x+xy < y+yx,$$
i.e., iff $-1 < x < y$. Hence, since $a < b + c$, and $b$, $c > 0$,
$$\frac{a}{1+a} < \frac{b+c}{1+b+c} = \frac{b}{1+b+c} + \frac{c}{1+b+c} < \frac{b}{1+b} + \frac{c}{1+c},$$
as required.