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 suppose $b,c\ge0$.
$$
\frac{b}{1 + b} + \frac{c}{1 + c} \ge \frac{a}{1 + a}
$$
$$
\Rightarrow \frac{b + c + 2bc}{1 + b + c +bc} \ge \frac{a}{1 + a}
$$
$$
\Rightarrow  b + c + 2bc + ab +ca + 2abc\ge a + ab + ca + abc
$$
$$
\Rightarrow b + c + 2bc + abc \ge a
$$
Add
$$
b + c \ge a \quad \textrm{and} \quad bc(a +2)\ge0
$$
And we are done.
