If $|f(0)|\geq r$, then $|f(z)|\geq (r-|z|)/(1-r|z|)$ for $|z|The following is an exercise IX.1.5 in Gamelin

Suppose that $f(z)$ is analytic and satisfies $|f(z)|\leq 1$ for $|z|<1$. Show that if $|f(0)|\geq r$, then $|f(z)|\geq (r-|z|)/(1-r|z|)$ for $|z|<r$.

Let $\varphi:\Bbb D\to\Bbb D$ be a Blaschke factor $\varphi(z) = {f(0)-z\over 1-\overline{f(0)}z}$. Then applying Schwarz lemma on $\varphi\circ f$, we get
\begin{align*}
& |(\varphi\circ f)(z)|\leq |z|\Rightarrow \left|{f(0)-f(z)\over 1-\overline{f(0)}f(z)}\right|\leq |z|.\\
& |f(0)|-|f(z)|\leq |f(0)-f(z)|\leq |z||1-\overline{f(0)}f(z)|\leq |z|(1+|f(0)||f(z)|).\\
& r(1-|z||f(z)|)\leq |f(0)|(1-|z||f(z)|)\leq |z|+|f(z)|.\\
& {r-|z|\over 1+r|z|}\leq |f(z)|.
\end{align*}
This is as far as I get. I wonder if there's a way to get the stated inequality.
 A: The inequality is correct, and can be proven with a small modification of your attempt. The “trick” is to multiply $f$ with a suitable constant such that $|1-\overline{f(0)}f(z)|$ can be estimated without applying the triangle inequality.
Fix $z_0 \in \Bbb D$ with $|z_0| < r$ and define the Blaschke factor $\varphi$ as
$$
\varphi(w) = \frac{f(0)-c w}{1-\overline{f(0)}c w}
$$
where $c \in \Bbb C$ is chosen such that $|c|=1$ and that $\overline{f(0)}cf(z_0)$ is real and non-negative. Then the Schwarz lemma can be applied to $\varphi \circ f$ and gives
$$
 \left| \frac{f(0)-c f(z)}{1-\overline{f(0)}c f(z)}\right| \le |z|
$$
for all $z \in \Bbb D$. In particular for $z = z_0$ we get
$$
|f(0) - cf(z_0)| \le  |z_0|\cdot|1-\overline{f(0)}c f(z_0)| = |z_0|\cdot (1-|f(0)||f(z_0)|)
$$
where the equality on the right follows from the choice of $c$. Then
$$ 
 |f(0)| - |f(z_0)| \le |f(0) - cf(z_0)|  \le  |z_0|\cdot (1-|f(0)||f(z_0)|)  
$$
and therefore
$$
 |f(z_0)| \ge \frac{|f(0)|-|z_0|}{1-|f(0)||z_0|}  \ge \frac{r-|z_0|}{1-r|z_0|} \, .
$$
