Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.

Let us prove the equivalent inequality
$$\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} > 0.$$
Since $\frac{n}{1+n} = 1 - \frac{1}{1+n}$, observe that
$$\begin{align*}
\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a}
&= 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+a}\\
&> 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+b + c + bc}\\
&= \left ( 1 - \frac{1}{1+b} \right ) \left ( 1 - \frac{1}{1+c} \right ) \\
&> 0.
\end{align*}$$
The first inequality used is
$$\frac{1}{1+a} > \frac{1}{1 + b + c + bc},$$
which is true since $a < b + c$ and $0 < bc$.
The final inequality holds since $b, c > 0$, so both terms in parentheses are positive.
