Given $|b-c|\leq a\leq b+c$ show that $\frac{a}{1+a}\leq\frac{b}{1+b}+\frac{c}{1+c}$ I am trying to show the following:

Given that $|b-c| \leq a \leq b+c$ show that:
$$\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$$

So far I have done the following:
$a/(1+a) \leq b/(1+b) + c/(1+c) \iff$
$1-1/(1+a) \leq 2 - 1/(1+b) -1/(1+c) \iff$
$1/(1+b) -1/(1+c) \leq 1 + 1/(1+a) \iff$
$ (2+b+c)/(1+b)(1+c) \leq (2+a)/(1+a) \iff$
$ (2+b+c)(1+a) \leq (2+a)(1+b)(1+c) \iff$
$ a \leq bc(2+a)$
I am not really sure how to proceed from here- any help or hints would be much appreciated.

I will assume that $a,b,c > 0$ because without such an assumption we cannot even have $1+b$ and $1+c$ in the denominator.
$f(x)=\frac x {1+x}$ is an increasing function of $x$ on $[0,\infty)$. Hence $\frac a {1+a} \leq \frac {b+c} {1+b+c} =\frac b {1+b+c}+\frac c {1+b+c} \leq \frac b {1+b} +\frac c {1+c} $.
