1.1 Lemma Le $f$ be an analytic in $D=D(0,1)$ and suppose that $f(0)=0$, $f'(0)=1$, and $\vert f(z)\vert \leq M$ for all $z$ in $D$. Then $M\geq 1$ and $f(D)\supset B(0, 1/6M)$.
Proof. Let $0<r<1$ and $f(z) = z + a_{2}z^{2}+ \ldots$; according to Cauchy's Estimate $\vert a_{n} \vert \leq M/r^{n}$. So $1=a_{1}\leq M$. If $\vert z\vert =(4M)^{-1}$ then
\begin{align*} \vert f(z)\vert &\geq \vert z\vert - \sum_{n=2}^{\infty}\vert a_{n}z^{n}\vert\\ &\geq (4M)^{-1} - \sum_{n=2}^{\infty} M(4M)^{-n}. \end{align*}
Why is $a_{1}\leq M$ ? The inequality give $a_{1}\leq M/r$ ...
Second, why is it $M$ in the sum ?
$|a_n| \leq M/r^{n}$ for every $r \in (0,1)$. Letting $r$ increase to $1$ we get $|a_n| \leq M$.