$ 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 $|v|$

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 $z \to 1$.
