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

This holds for complex numbers as well, and the expression is called a Blaschke factor. See https://en.wikipedia.org/wiki/Blaschke_product. — Reiner Martin, Nov 22 '17 at 21:23
Also, this seems to be a duplicate: https://math.stackexchange.com/questions/506058/show-that-a-complex-expression-is-smaller-than-one and https://math.stackexchange.com/questions/1562930/show-that-blaschke-factors-satisfy-inequality. — Reiner Martin, Nov 22 '17 at 21:25

score 2 · Answer 1 · answered Nov 22 '17 at 21:26

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.

answered Nov 22 '17 at 21:26

Michael Rozenberg

score 1 · Answer 2 · answered Nov 22 '17 at 21:22

1

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

answered Nov 22 '17 at 21:22

nonuser

2 Answers2