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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

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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 $ab+1=(a-1)(b-1)+(a+b)>a+b=|a+b|$.

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  • $\begingroup$ Thank you, that makes sense $\endgroup$ – user523384 Jun 10 at 5:16

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