upper bound for a ratio of complex numbers Suppose $a,b,c,d$ are complex numbers  such that  $|a|=|b|=1$ and $|c|,|d|\geq R>1$  and I believe that   the following inequality is true;
$ \left|\frac{(a-c)+(a-d)}{2(b-c)(b-d)} \right|\leq \frac{1}{1+R}$ whenever  $|(b-c)(b-d)|\geq |(e^{i\theta}-c)(e^{i\theta}-d)|$ for all real values of $\theta.$  How  to validate this? Kindly suggest with hint/idea..
 A: Wlog assume $c=R, d=Re^{i\theta}, 0 \le \theta \le \pi$ using the symmetries of the problem (rotation and reflection in the $x$ axis).
Then since $|c+d|=2R\cos \theta/2$ one immediately sees that the maximum of the numerator is $2R\cos \theta/2+2$ by choosing $a$ opposite to $c+d$ so $|c+d-2a|=|c+d|+2|a|=|c+d|+2$
But one can see that $b$ is the point on the circle of radius $1$ where the bisector median of $cd$ intersects it farthest (note that since by definition $b$ maximizes the denominator, we do not need to prove that the $b$ above is the good one, just prove the inequality for it, as of course it then holds for any putative $b_1$ giving a higher denominator, so I am not going to prove the above statement about the chosen $b$, just prove the inequality as that solves the problem),
If we consider the triangle $BOC$ say (where $B$ corresponds to $b$, $O$ is the origin and $C$ corresponds to $c$), if follows that since the angle $DOC$ is $\theta$ by definition, then the angle $BOC$ is $\pi-\theta/2$ so the denominator which is $BC^2$ is $1+R^2+2R\cos \theta/2$
So we need to prove:
$$\frac{2R\cos \theta/2+2}{2(1+R^2+2R\cos \theta/2)} \le \frac{1}{1+R}$$
But this is equivalent to
$$1+R+R\cos \theta/2+R^2\cos \theta/2 \le 1+R^2+2R\cos \theta/2$$ or to $$R(1-\cos \theta/2) \le R^2(1-\cos \theta/2)$$
which is obvious since $R > 1$
We get equality precisely when $\theta=0$ so $c=d=R$ (or $c=d$ in general before rotation)
