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
$$\frac{|i(Re^{i\theta})^\lambda|}{|1+Re^{i\theta}|}\leq\frac{R^\lambda}{R-1}$$