$\frac{|x+y|}{|1+xy|}<1$ for $|x|<1$ and $|y|<1$

Question

If we consider $x,y\in\mathbb{R}$ with $|x|<1$ and $|y|<1$. How do we get to $$\frac{|x+y|}{|1+xy|}<1?$$

Answer 1

$$|x+y|<|1+xy|\;\;\;\;/^2$$ $$x^2+2xy+y^2<1+2xy+x^2y^2$$ $$ 0<(x^2-1)(y^2-1)$$ $$ 0<(|x|^2-1)(|y|^2-1)$$

Answer 2

We need to, prove that $$-1<\frac{x+y}{1+xy}<1,$$ which is $$(1+x)(1+y)>0$$ and $$(1-x)(1-y)>0,$$ which is obvious.