I need help proving this inequality. I managed to go through some steps but I'm not sure if and how I can continue with my way:
What needs to be proved:
$$(|a+b|)/(1+|a+b|)≤(|a|)/(1+|a|)+|b|/(1+|b|)$$
Here is my progress: $$(|a|)/(1+|a|)+(|b|)/(1+|b|)\ge \\ (|a|)/(1+|a|+|b|)+(|b|)/(1+|a|+|b|)= \\ (|a|+ |b|)/(1+|a|+ |b| )\ge \\ (|a+b|)/(1+|a|+ |b|)$$
What I still need to prove but can't manage:
$$ (|a+b|)/(1+|a|+ |b|)≥(|a+b|)/(1+|a+b|)$$
Can it be done this way or should I try a different way?
The function $x\in [0,\infty)\mapsto \frac{x}{1+x}$ (by calculating its derivative or by observing what Bernard pointed out). So as $|a+b|\le |a|+|b|$, we have:
$$\frac{|a+b|}{1+|a+b|} \le \frac{|a|+|b|}{1+|a|+|b|} \\ = \frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|} \le \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$$