Give an example when the equality sign holds.
My approach is like this, for $w\in U$, define $\phi_w=\frac{z-w}{z-\bar{w}}.$ Then each $\phi_w$ is conformal map from $U$ to $D$ sending $w$ to $0.$ Then, define $h(z)=\phi_{f(i)}(f(\phi_i^{-1}(z)))$, it is a holomorphic map from $D$ to $D$ and $h(0)=0$. Schwarz Lemma tells that$$1\geq|h'(0)|=|\frac{1}{f(i)-\overline{f(i)}}||f'(i)||2i|.$$ We then have $|Im f(i)|\geq|f'(i)|$. I then don't know how to construct the case where the equality could hold. Since equality holds mean $|Im f|=|f|$, i.e. $f$ maps $U$ to the upper imaginary axis, contradicting open mapping theorem. Could you please point out my mistakes? That should be $|f(i)|=|Im f(i)|$, This is at $i$ only, hence not contradicting open mapping theorem.