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I have to show that the inequality below is true, i tried some thing but got stuck, i tried to eliminate the absolute value $-1<\frac{x-y}{1-xy}<1$ and then solve for $x$ and $y$ with no luck...i do not want the answer to this problem but at least a method for solving this kind of exercises.

$$x,y\in\Bbb R\ , |x|<1,|y|<1 \ \ \text{show that} \ \ \bigg|\frac{x-y}{1-xy}\bigg| < 1.$$

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2 Answers 2

up vote 3 down vote accepted

Observe:

$|x - y|^2 < |1 - xy|^2 \Leftrightarrow (1 - |x|^2)(1 - |y|^2) > 0$

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I do not understand why the last equality proves everything could you elaborate a bit more? –  phi Nov 20 '12 at 18:37
    
Since both numbers $\,x,y\,$ are less than one in absolute value, their squares are less than one, too... –  DonAntonio Nov 20 '12 at 18:39

$$\left|\frac{x-y}{1-xy}\right|<1\Longleftrightarrow |x-y|<|1-xy|\Longleftrightarrow x^2-2xy+y^2<1-2xy+x^2y^2\Longleftrightarrow$$

$$\Longleftrightarrow (x^2-1)(y^2-1)>0$$

and since the last inequality above is trivially true we're done

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