Let , $f$ be non-constant holomorphic function in a nbd. of $\bar{\mathbb D}$ with $f(0)=a_0$. Let , $\displaystyle M=\max_{z\in \mathbb D}|f(z)$. Let, $\lambda \in \mathbb D$ and $f(\lambda)=0$. Prove that $|\lambda| \ge \frac{|a_0|}{M}$.
First consider the function , $$g(z)=f\left(\frac{z+\lambda}{1+\bar{\lambda}z}\right).$$Then $g$ is analytic in $\mathbb D$ onto $\mathbb D$. Also , $g(0)=f(\lambda)=0$. So by Schwarz lemma , $|g(z)|\le|z|$ for all $z\in \mathbb D$. In particular for $z=-\lambda$ , $|g(-\lambda)|=|\lambda|\implies|\lambda|\ge |a_0|$. But I am unable to do the proof where $M$ is the fact.
Schwarz Lemma provides for your $g$ that $\,\lvert g(z)\rvert\le M\lvert z\rvert$. (Apply it on $g/M$, not on $g$.) Hence $$ \lvert a_0\rvert=\lvert \,f(0)\rvert=\lvert \,g(-\lambda)\rvert\le M\lvert \lambda\rvert. $$