Let $f$ be an anylytic function on the unid disk $|z|<1$. It is well known that 
$$\left (\int\_0^{2\pi}f(e^{i\theta})d \theta \right)^2 \geq 4\pi \iint\_{|z|<1} |f(r e^{i\theta})|^2r dr d \theta.$$


I wonder if the constant $4\pi$ cound be improved on an annulus $a<z<1$. More precisely, does the following inequality 


$$\left (\int\_0^{2\pi}|f(e^{i\theta})|d \theta +\int\_0^{2\pi}|f(ae^{i\theta}|)d \theta \right)^2 \geq C \iint\_{a<|z|<1} |f(r e^{i\theta})|^2r dr d \theta$$


hold for some constant $C(a)> 4\pi$ independent of $f$? Can $C(a)$ be computed in terms of $a$? This seems to be a classical problem but I could not find a reference, and was not able to prove it after trying for a couple of days. 



Isoperimetric inequality for analytic functions on an annulus
You can probably prove that 
$$ \Big( \int\_\mathbb{A\_r} |f(z)|^2 \frac{dxdy}{\pi(1-r^2)} \Big)^{1/2} \leq \int\_{ \mathbb{T}} |f(e^{i\theta})| \frac{d\theta}{2\pi}+\int\_{ \mathbb{T\_r}} |f(re^{i\theta})| \frac{d\theta}{2\pi r}, \,\,\, \forall f\in H^1(\mathbb{A\_r}). $$
Where $\mathbb{A\_r} $ is the annulus of internal radius $r$ with boundary $\mathbb{T} \cup \mathbb{T\_r}$, and $H^1(\mathbb{A\_r})$ is the Hardy space on the annulus, by following step by step the proof of Vukotic in "The isoperimetric inequality and a theorem of Hardy and Littlewood, American mathematical monthly". The only tools you need is the Parseval formula and the inner outer factorization for the Hardy space in an annulus which you can find in "Sarason, The Hp spaces of an annulus, Memoirs or the AMS". If you have problems recustructing the proof, maybe I can help. 


