I just need to show that :
$\int_0^{2\pi}\left|{\frac{i(Re^{i\theta})^\lambda}{1+Re^{i\theta}}}\right| d\theta \le \int_0^{2\pi} \frac{R^\lambda}{R-1}d\theta : 0 < \lambda <1 , R>1$ Is there some trivial geometrical argument I don't see?
I just need to show that :
$\int_0^{2\pi}\left|{\frac{i(Re^{i\theta})^\lambda}{1+Re^{i\theta}}}\right| d\theta \le \int_0^{2\pi} \frac{R^\lambda}{R-1}d\theta : 0 < \lambda <1 , R>1$ Is there some trivial geometrical argument I don't see?
You just have to use the following inequality, valid for complex numbers $a, b \in \mathbb{C}$
$|a| - |b| \leq |a \pm b|$
which in this case applied to $a = Re^{i\theta}$ and $b = 1$ becomes
$ |Re^{i\theta}| - |1| \leq |Re^{i\theta} + 1| $
so this implies that
$ \frac{1}{|Re^{i\theta} + 1|} \leq \frac{1}{|Re^{i\theta}| - |1|} = \frac{1}{R - 1} $
Then using this, proving the inequality with the integrals should be straightforward to you.
$|i(Re^{i\theta})^\lambda| = R^\lambda$
$|1+Re^{i\theta}| \geq |Re^{i\theta}| - |1| = R - 1\rightarrow \frac{1}{|1+Re^{i\theta}|}\leq\frac{1}{R-1}$
then
$\frac{|i(Re^{i\theta})^\lambda|}{|1+Re^{i\theta}|}\leq\frac{R^\lambda}{R-1}$