A holomorphic function inequality. Is it right?

Question

I have proved using Green's theorem the following equality:$$\frac{1}{2\pi i}\left(\frac{1}{\zeta}-\bar{\zeta}\right)\int_{\partial \mathbb{D}}f(z+\zeta)\log\left(\frac{z}{z-\zeta}\right)dz=f(\zeta)+\frac{\zeta}{\pi}\iint_{\mathbb{D}}\frac{\bar{z}-\bar{\zeta}}{1-\zeta \bar{z}}f(z+\zeta)dxdy $$ For some $\zeta$ inside the disc and for $f:\Omega\to \mathbb{C}$ , $\mathbb{D} \subset \Omega$. Now I would like to estimate the integral in the left side. I consider a $\zeta$ such that $0<|\zeta|<1$ Now i would like to estimate $f(\zeta)$. After calculating the contour integral with residues, what i get is: $$f(\zeta)=\zeta f(2\zeta)\left(\frac{1-|\zeta|^2}{\zeta}\right)+\frac{\zeta}{\pi}\iint_{\mathbb{D}}\frac{\bar{z}-\bar{\zeta}}{1-\zeta \bar{z}}f(z+\zeta)dxdy$$ and after using the triangle inequality I get: $$|f(\zeta)|\le (1-|\zeta|^2)|f(2\zeta)|+\frac{|\zeta|}{\pi}\iint_{\mathbb{D}}|f(z+\zeta)|dxdy.$$ If $f$ is univalent then we substitute $f(z+\zeta) $ with $(f^{'}(z+\zeta))^2 $ one has a lower bound for the area of this function inside the unit disc. $$\frac{|f^{'}(z)|^2-(1-|z|^2)|f^{'}(2z)|^2}{|z|}\le \mathrm{Αrea} f(\mathbb{D})$$ Are my estimations meaningful ?