Let $f$ be a holomorphic function on $D(0,1)$ such that $|f(z)|<1$ for all $z\in D(0,1)$. I have obtained $$ \frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)|\leq \frac{|f(0)|+|z|}{1-|f(0)||z|}. $$
Is it true
$$ \frac{|f(0)|-|z|}{1-|f(0)||z|}\leq |f(z)|\leq \frac{|f(0)|+|z|}{1+|f(0)||z|} $$ for all $z\in D(0,1)$? Any counter examples?
This is indeed true. One way that immediately comes to mind is to use Schwarz Lemma. But first we need to ensure that $f(0) = 0$. Here we can use an automorphism of the unit disk such that if $f(0) = a$, then both $\phi_a \circ f (0) = \tilde{\phi}_a \circ f(0) = 0 $ where $$ \phi_a (z) = \frac{a-z}{1-\bar{a}z} ; \qquad \tilde{\phi}_a(z) = \frac{z-a}{1-\bar{a}z}$$
Now using Schwarz lemma on each of the composite functions will give the two sided inequality. for all $z \in D(0,1)$.