Proving a bound for analytic functions on the disc

Question

Let $f$ be holomorphic on $|z| < 2$

I want to show that $\max_{|z| = 1} | f(x) - {1 \over z}| \geq 1 $

I think the idea is to apply a Mobius transform to $f(x) - {1 \over z}$ and then apply either maximum modulus or the Schwarz principle.

Answer 1

Integrating the function $f(z)-\dfrac1z$ over the path $|z|=1$ we get $$ \int_{|z|=1}\left(f(z)-\frac1z\right)dz=-2\pi i. $$ On the other hand, $$ \left|\int_{|z|=1}\left(f(z)-\frac1z\right)dz\right|\le\int_{|z|=1}\left|f(z)-\frac1z\right|dz\le2\pi\max_{|z|=1}\left|f(z)-\frac1z\right|. $$ Hence, $\max_{|z|=1}\left|f(z)-\dfrac1z\right|\ge1$.