If $0<y<1$, and $-1<x<1$, then prove that $\left |\frac{x(1-y)}{1+yx}\right| < 1$

Question

If $0 < y < 1$ and $-1 < x<1$, then prove that $$\left|\frac{x(1-y)}{1+yx}\right| < 1$$

Answer 1

to prove $|\frac{x(1-y)}{1+yx}|<1$ it suffices to show $|1+yx|>|x(1-y)|$.

If $x\geq 0$ then $|1+yx|=1+yx>x>x|1-y|=|x(1-y)|$

if $x<0$ then $|1+yx|=1+yx>1-y>|x||1-y|=|x(1-y)|$

Answer 2

$$ \bigg|\frac{x(1-y)}{(1+yx)}\bigg|=\frac{|x|(1-y)}{1+yx}. $$

We have two cases.

If $x\geq 0$, then the above becomes

$$ \frac{x(1-y)}{1+xy}. $$

Note that $x(1-y)<x<1$, while $1+xy>1$. Thus, the inequality follows in this case.

If instead $x<0$, then the inequality becomes

$$ \frac{-x(1-y)}{1+xy}=\frac{-x+xy}{1+xy}. $$

Note that $-x<1$ and, hence, $-x+xy<1+xy$ so the inequality follows.

Answer 3

This will hold true iff $$x^2(1-y)^2<(1+xy)^2$$

$$\iff0>x^2-2x^2y-2xy-1=(x+1)(x-1-2xy)$$

$$\iff0>x-1-2xy\iff1>x(1-2y)$$

WLOG let $y=\sin^2A,x=\cos B$

$x(1-2y)=\cos B\cos2A$ which is $<1$

OR

as $0<y<1\iff0>-2y>-2\iff1>1-2y>-1$ and we have $-1<x<1$