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.
 A: Function $f(x)=\frac{x}{1+x}$ is increasing (it is easy to check first derivative). Therefore $f(a)<f(b+c)$, for $a<b+c$. Because of that, we have $$\frac{a}{1+a}<\frac{b+c}{1+b+c}=\frac{b}{1+b+c}+\frac{c}{1+b+c},$$ from where we get (since $a,b,c>0$):
$$\frac{a}{1+a}<\frac{b}{1+b}+\frac{c}{1+c}.$$
A: 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.
A: By C-S
$$\frac{b}{1+b}+\frac{c}{c+1}=\frac{b^2}{b^2+b}+\frac{c^2}{c^2+c}\geq\frac{(b+c)^2}{b^2+c^2+b+c}\geq\frac{(b+c)^2}{(b+c)^2+b+c}=$$
$$=\frac{b+c}{b+c+1}=1-\frac{1}{b+c+1}>1-\frac{1}{a+1}=\frac{a}{a+1}.$$
Done!
A: $$\frac{b}{1+b} + \frac{c}{1+c}>\frac{a}{a+1}$$
$$\frac{b+c+2bc}{1+b+c+bc}>\frac{a}{a+1}$$
Since $a,b,c>0$, 
$$ab+ac+2abc+b+c+2bc>a+ab+ac+abc$$
$$abc+2bc+b+c>a$$
