$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w  \in  \mathbb{C}  $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Well, i could only proof that
$ u=\left(\frac{z + w}{R^2 + zw}\right) $ , $ v = \left(\frac{z - w}{R^2 -  zw}\right) \in \mathbb{R} $
By showing that
$ u = \overline{u}$ and $ v = \overline{v}$
Which leads to
$ u^2 + v^2 = |u|^2 + |v|^2 $
But i can't find a way to compute  $|u|$ or $|v|$
 A: We must assume that $zw \ne \pm R^2$ because the left-hand side is undefined otherwise. Then we can assume that $R=1$ because of the homogeneity of the inequality, this simplifies the calculation a bit.
You already noticed that the terms on the left-hand side are real numbers, so that
$$
\left(\frac{z + w}{1 + zw}\right)^2 + 
\left(\frac{z - w}{1 - zw}\right)^2 =
\left|\frac{z + w}{1 + zw}\right|^2 + 
\left|\frac{z - w}{1 - zw}\right|^2 \, .
$$
Now one can apply the identity $|a+b|^2 = |a|^2 + |b|^2 + 2 \operatorname{Re}(\bar a b)$ multiple times.
Our expression becomes
$$
\frac{2+2\operatorname{Re}(\bar z w)}{2+2\operatorname{Re}( z w)} +
\frac{2-2\operatorname{Re}(\bar z w)}{2-2\operatorname{Re}( z w)} \, .
$$
To simplify the calculation further we can set $s = \operatorname{Re}(\bar z w)$ and $t = \operatorname{Re}( z w)$. Then $|s|\le 1$ and $|t| < 1$ and our expression is equal to
$$
 \frac{1+s}{1+t} + \frac{1-s}{1-t} = \frac{2-2st}{1-t^2} = 1 + \frac{t^2-2st+1}{1-t^2} > 1
$$
because $t^2-2st+1 = (t-s)^2 + (1-s^2) > 0$.
The inequality is strict and sharp: For $z=w \ne \pm 1$ we have
$$
\left|\frac{2z}{1 + z^2}\right|^2 + 0^2 = \frac{4}{|1+z^2|^2} \to 1
$$
for $z \to 1$.
A: 
From the picture, if $\arg{(w)}=a$ and $\arg{(z)}=b$, then
$$
\begin{aligned}
|z+w|&=2R\left|\cos{\left(\frac{a-b}{2}\right)}\right|\\
|z-w|&=2R\left|\sin{\left(\frac{a-b}{2}\right)}\right|\\
\\
|R^{2}+zw|&=2R^{2}\left|\cos{\left(\frac{a+b}{2}\right)}\right|\\
|R^{2}-zw|&=2R^{2}\left|\sin{\left(\frac{a+b}{2}\right)}\right|
\\
\\
\left(\frac{z+w}{R^{2}+zw}\right)^{2}+\left(\frac{z-w}{R^{2}-zw}\right)^{2}&=\frac{1}{R^{2}}\frac{\cos^{2}{\left(\frac{a-b}{2}\right)}}{\cos^{2}{\left(\frac{a+b}{2}\right)}}+\frac{1}{R^{2}}\frac{\sin^{2}{\left(\frac{a-b}{2}\right)}}{\sin^{2}{\left(\frac{a+b}{2}\right)}}
\end{aligned}
$$
Which is always greater than $\frac{1}{R^{2}}$
A: Written as $\left(\frac{z+w}{|z||w|+zw}\right)^2+\left(\frac{z-w}{|z||w|-zw}\right)^2\ge\frac1{|z||w|}$, the inequality scales nicely, so we can assume wlog that $R=|z|=|w|=1$.
Set $z=e^{i\alpha}$ and $w=e^{i\beta}$. If $(zw)^2\ne1$, then $\cos^2(\alpha+\beta)\ne1$. Furthermore,
$$
\begin{align}
&\left(\frac{z+w}{1+zw}\right)^2+\left(\frac{z-w}{1-zw}\right)^2\\
&=2\,\frac{z^2+w^2+z^4w^2+z^2w^4-4z^2w^2}{1-2z^2w^2+z^4w^4}\tag1\\
&=2\,\frac{\bar{w}^2+\bar{z}^2+z^2+w^2-4}{\bar{z}^2\bar{w}^2-2+z^2w^2}\tag2\\
&=2\,\frac{2\cos(2\alpha)+2\cos(2\beta)-4}{2\cos(2\alpha+2\beta)-2}\tag3\\
&=2\,\frac{1-\cos(\alpha+\beta)\cos(\alpha-\beta)}{1-\cos^2(\alpha+\beta)}\tag4\\[6pt]
&\gt1\tag5
\end{align}
$$
Explanation:
$(1)$: expand and combine
$(2)$: multiply by $\frac{\bar{z}^2\bar{w}^2}{\bar{z}^2\bar{w}^2}$
$(3)$: $z^2+\bar{z}^2=2\cos(2\alpha)$, $w^2+\bar{w}^2=2\cos(2\beta)$
$\phantom{\text{(3):}}$ $z^2w^2+\bar{z}^2\bar{w}^2=2\cos(2\alpha+2\beta)$
$(4)$: $\cos(2\alpha)+\cos(2\beta)=2\cos(\alpha+\beta)\cos(\alpha-\beta)$
$\phantom{\text{(4):}}$ $\cos(2\alpha+2\beta)=2\cos^2(\alpha+\beta)-1$
$(5)$: apply $(6)$ below
Since $\frac{1-xy}{1-x^2}$ is monotonic in $y$ for each $x\in(-1,1)$, we have for $y\in[-1,1]$
$$
\frac12\lt\frac1{1+|x|}\le\frac{1-xy}{1-x^2}\le\frac1{1-|x|}\tag6
$$
