The derivative of a holomorphic function on the boundary of the unit circle Let $f$ be holomorphic on $D(0,1)\cup \{1\}$, and $f(0)=0$, $f(1)=1$, $f(D(0,1))\subset D(0,1)$. Prove that $|f'(1)|\geq 1$.
I have no idea. Maybe $f'(\xi)=[f(1)-f(0)]/(1-0)$ for some $\xi \in (0,1)$, and the maximal modules principle applies? This is not exact.

It is easier to argue by contradiction. We are given that $f'(1)$ exists. Suppose $|f'(1)|<1$. Pick $c$ so that $|f'(1)|<c<1$. There  is a neighborhood of $1$ in which $$|f(z)-f(1)|\le c|z-1|$$ By the reverse triangle $$|f(z)|\ge |f(1)|-c|z-1| = 1-c|z-1| \tag{1}$$
On the other hand, by the Schwarz lemma 
$$|f(z)|\le |z|  \tag{2}$$
From (1) and (2) we get 
$$1-|z|\le c|z-1|\tag{3}$$
which is impossible when $z\in (0,1)$.
