Prove $|(1+|u|^2)v-(1+|v|^2)u|>|u\bar{v}-\bar{u}v|$ Show that for $u,v \in \mathbb{C}$ with $|u|<1, |v|<1$, and $\bar{u}v\neq u\bar{v}$, we always have
$$\left|\left(1+|u|^2\right)v-\left(1+|v|^2\right)u\right|>\left|u\bar{v}-\bar{u}v\right|.$$

Divide both sides of the inequality we want to prove by $(1+|u|^2)(1+|v|^2)$, we find
it is equivalent to following statement
$$| y - x | \stackrel{?}{>} |x\bar{y} - y\bar{x}|
\quad\text{ where }\quad x = \frac{u}{1+|u|^2}\quad\text{ and }\quad y = \frac{v}{1+|v|^2}
$$
Notice 
$$\begin{align}
& \bar{u} v \ne u\bar{v} \implies x \ne y \implies |x-y| > 0\\
\text{ and }\quad & \;|v| < 1\; \implies |y| = \frac{|v|}{1+|v|^2} = \frac{1}{|v|+|v|^{-1}} < \frac12
\end{align}$$
We have
$$|x\bar{y} - y\bar{x}| = |(x - y)\bar{y} + y(\bar{y} - \bar{x})|
\le |x-y||\bar{y}| + |y||\bar{y}-\bar{x}| = 2|x-y||y| < |x-y|$$
This means the inequality we want to prove is indeed true.
