# Prove that $|\frac{a+b}{1+ab}| < 1$ given that $|a|<1$ and $|b|<1$

Can I please get a hint/solution for why, if true,

$$\bigg{|}\frac{a+b}{1+ab}\bigg{|}<1$$

given that $$|a|<1$$ and $$|b|<1$$, where $$a,b \in \mathbb{R}$$.

I've tried the usual things like deleting something from the top/bottom or triangle inequality but I have no idea how to do it. Maybe I'm missing something obvious?

Thanks

It suffices to show $$ab+1 > |a+b|$$ since $$ab+1>0$$.
Also, we may assume $$a+b\geq0$$.
(If not, substitute $$\alpha = -a$$ and $$\beta = -b$$ and consider the statement for $$\alpha$$ and $$\beta$$.)
Now $$ab+1=(a-1)(b-1)+(a+b)>a+b=|a+b|$$.