Inequality $\lvert(1+\lvert u\rvert^2) v-(1+\lvert v\rvert^2)u\rvert > \lvert u\bar{v}-\bar{u}v\rvert$ for complex $u,v$ This is exercise 4.(b) from Theory of Complex Functions, GTM 122, page 17. I was able to solve exercise 4.(a), but I don't see how this could help solving 4.(b):

a) Show that from $(1+\lvert v\rvert^2) u=(1+\lvert u\rvert^2)v,\,u,v\in\mathbb{C}$, it follows that either $u=v$ or $\bar{u}v=1$.
b) Show that for $u,v\in\mathbb{C}$ with $\lvert u\rvert<1, \lvert v\rvert<1$ and $\bar{u}v\ne u\bar{v}$, we always have $$\lvert(1+\lvert u\rvert^2) v-(1+\lvert v\rvert^2)u\rvert > \lvert u\bar{v}-\bar{u}v\rvert$$

For a), I took modulus of both sides and get either $\lvert u\rvert=\lvert v\rvert$ or $\lvert u\rvert \lvert v\rvert=1$, where the conclusion follows easily. For b), I tried substituting $u=x+iy$ and $v=p+iq$ but the resulting inequality is too complicated and it's difficult to see where the condition $\lvert u\bar{v}-\bar{u}v\rvert=2\lvert xq-yp\rvert\ne 0$ is used. Any advice?
 A: Notice that both sides remain the same if we substitute $u \to cu, v \to cv$ for $|c| = 1$. Thus without loss of generality, we can assume $u$ is real. Then
$$\begin{align*}Left&= |(1+|u|^2)v - (1+|v|^2)u| \\
&= |\left((1+|u|^2) \mathrm{Re} (v) - (1+|v|^2)u\right) + \left((1+|u|^2) \mathrm{Im}(v)\right) i|\\
&\text{Since norm of a complex number is at least absolute value of its imaginary part,} \\
&\ge |(1+|u|^2) \mathrm{Im} (v)| \\
&= (1+|u|^2) |\mathrm{Im}(v)| \\
&\ge 2|u| |\mathrm{Im}(v)| \\
&= |u| |\overline{v} - v| \\
&= |u\overline{v} - \overline{u}v| \text{ (since $u$ is real)} \\
&= Right
\end{align*}
$$
A: For Part (a), we note that $\big(1+|v|^2\big)\,u=\big(1+|u|^2\big)\,v$ implies  $\big(1+|v|^2\big)\,|u|=\big(1+|u|^2\big)\,|v|$.  That is,
$$\big(|u|-|v|\big)\,\big(1-|u|\,|v|\big)=0\,.$$
If $|u|=|v|$, then we obtain $u=v$.  If $u\neq v$, then $|u|\,|v|=1$, whence $u\neq 0$ and $v=\dfrac{1}{|u|^2}\,u$.  Ergo,
$$\bar{u}\,v=\bar{u}\,\left(\frac{1}{|u|^2}\,u\right)=\frac{|u|^2}{|u|^2}=1\,.$$
For Part (b), we shall prove that
$$\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|\geq \big|u\,\bar{v}-\bar{u}\,v\big|\tag{*}$$
for all $u,v\in\mathbb{C}$, and the equality occurs if and only if $u=v$.  If $u=0$ or $v=0$, then (*) trivially holds.  From now on, we assume that $u\neq 0$ and $v\neq 0$.
Observe that
$$\begin{align}\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&=\Big(\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big)\,\Big(\big(1+|v|^2\big)\,\bar{u}-\big(1+|u|^2\big)\,\bar{v}\Big)
\\
&=\big(1+|v|^2\big)^2\,|u|^2+\big(1+|u|^2\big)^2\,|v|^2-2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,\text{Re}\big(u\,\bar{v}\big)\,.
\end{align}$$
Using the AM-GM Inequality on $\big(1+|v|^2\big)^2\,|u|^2+\big(1+|u|^2\big)^2\,|v|^2$, we obtain
\begin{align}\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&\geq2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,|u|\,|v|-2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,\text{Re}\big(u\,\bar{v}\big)\\
&= 2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,\Big(|u|\,|v|-\text{Re}\big(u\,\bar{v}\big)\Big)\,.
\end{align}
Because $\big(\text{Im}(z)\big)^2=z^2-\big(\text{Re}(z)\big)^2$ for every $z\in\mathbb{C}$, we get
\begin{align}\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&\geq2\,\left(\frac{\big(1+|u|^2\big)\,\big(1+|v|^2\big)}{|u|\,|v|+\text{Re}\big(u\,\bar{v}\big)}\right)\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2
\\
&\geq 2\,\left(\frac{\big(1+|u|^2\big)\,\big(1+|v|^2\big)}{2\,|u|\,|v|}\right)\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2\,,
\end{align}
where we have applied the trivial inequality $\text{Re}(z)\leq |z|$ for each $z\in\mathbb{C}$.  Finally, using the AM-GM Inequality $t+\dfrac1t\geq 2$ for all $t>0$, we get
\begin{align}
\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&\geq\left(|u|+\frac{1}{|u|}\right)\,\left(|v|+\frac{1}{|v|}\right)\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2
\\
&\geq 4\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2=\big|u\,\bar{v}-\bar{u}\,v\big|^2\,,\end{align}
and the claim is proven.  (The equality case is easy to verify.)
