Prove that a function which is analytic in unit circle and satisfies $f(0)=1$ and $f(z) \ge 1 + |z|^ 2$ doesn't exist

Question

Let $G=\{ z \mid |z| < 1\} $. Show there are not function such that $f \in A(G)$ and $f(0)=1$ and $|f(z)| \ge 1 + |z| ^ 2$

So I assume by contradiction that there is such function $f$. I thought to start using Cauchy's formula and I get $$ \int_{|z|= 1} \frac{f(z)}{z} \,\mathrm dz= f(0) = 1$$

It seems that I should somehow use the maximum principle, but since here $f$ is greater than something, I am not really sure how.

Help would be appreciated

Answer 1

Suppose there is such an $f.$ Then the inequality $|f(z)|\ge 1+|z|^2$ implies $f$ is never $0$ in $G.$ Thus $1/f\in A(G).$ Now apply the maximum principle to see

$$1=\left |\frac{1}{f(0)}\right| \le \max_{|z|=1/2}\left |\frac{1}{f(z)}\right |$$ $$ = \max_{|z|=1/2}\frac{1}{|f(z)|} \le \frac{1}{1+(1/2)^2} = \frac{4}{5},$$

a contradiction.

Answer 2

This follows from the open mapping theorem. If such an $f$ existed, its image would have to contain an open disk around $1$ (since $f(0)=1$) and thus, in particular, a point $w$ with $|w| < 1$; but $|f(z)|\ge 1+|z|^2 \ge 1$ for all $z\in G$.