Estimating the modulus of a ratio For any $x,y> 0, x\leq y, xy\geq 1,$ and real values of $\theta,$ we can establish easily the inequality
$$\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|\leq \sqrt{\frac{x}{y}}, $$ for all $y$ and $\theta$ for which  $1-ye^{i\theta}\neq 0.$ But my idea is to relax the condition $xy\geq 1.$ Can we have a bound for $\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|$ for any $x,y\geq 0, x\leq y.$
 A: Due to the symmetry, it suffices to consider $0\leq\theta\leq \pi.$
Denote $O,A,X,Y$ the points corresponding respectively to the numbers $0,\;1,\;xe^{i\theta},\;ye^{i\theta}.$ 
Assume $0\neq\theta\neq \pi.$
With the use of the law of cosines in $\triangle OAX$ and $\triangle OAY$ we get
$$|1-xe^{i\theta}|^2=1+x^2-2x\cos\theta,$$ 
and
$$|1-ye^{i\theta}|^2=1+y^2-2y\cos\theta.$$


*

*If $xy<1,$ the function $g(\theta)=\frac{1+x^2-2x\cos\theta}{1+y^2-2y\cos\theta}$ is decreasing since $g'(\theta)=\frac{2\sin\theta\,(y-x)(xy-1)}{(y^2+1-2y\cos\theta)^2}.$ Thus $g(\pi)<g(\theta)<g(0).$ This gives the estimate $$\frac{1+x}{1+y}<\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|<\frac{|1-x|}{|1-y|}.$$

*Similarly, if $xy>1,$ we have $$\frac{1+x}{1+y}>\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|>\frac{|1-x|}{|1-y|}.$$

*A straightforward computation shows that for $xy=1$ we have $g(\theta)=x^2=\frac{1}{y^2},$ which corresponds to $$\frac{1+x}{1+y}=\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|=\frac{|1-x|}{|1-y|}.$$
If $A,X,Y$ are aligned, the above estimates hold (continuity of $g$ on $\mathbb{R}).$ 
Note
In the case $xy\geq1$ it holds (easy to verify) $$\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|\leq\frac{1+x}{1+y}\leq\sqrt{\frac{x}{y}},$$ thus we have a better upper-estimate. 
In notation of arithmetic and geometric means, we have 
$$\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|\leq\frac{AM(x,1)}{AM(y,1)}\leq\frac{GM(x,1)}{GM(y,1)}\quad \text{if}\quad xy\geq 1,$$ and
$$\left|\frac{1-xe^{i\theta}}{1-ye^{i\theta}}\right|>\frac{AM(x,1)}{AM(y,1)}>\frac{GM(x,1)}{GM(y,1)}\quad \text{if}\quad xy< 1.$$

