Let $f(z)$ be analytic in the unit disc,

Let $f(z)$ be analytic in the unit disc, $Rf(z) \ge 1$, and $f(0)>0$. Show that $$f(0) \cdot \frac{1-|z|}{1+|z|} \le |f(z)| \le f(0) \cdot \frac{1+|z|}{1-|z|} \quad z \in D$$

This looks like something I would have to prove using cases. I tired to throw some numbers in there to see what was going on but I am still not seeing how to prove this. I think I am missing or forgetting a concept to use.

I have attempted the Schwartz lemma as suggested below and assumed $f(0)=1$ I am not sure if this is the right path to take?

• Try Schwartz formula and harnacks inequality – user133100 Apr 12 '16 at 23:46

The linear fraction $g(w)=\frac{1-w}{1+w}$ maps the right half plane to the unit disk such that $g(1)=0$.
By applying Schwarz's lemma to the function $w\mapsto g(f(w))$, we have $|g(f(z))|\le |z|$.
The function $g^{-1}(w)=\frac{1-w}{1+w}$ maps the disk $|w|\le|z|$ to the closed disk with diameter $\left(\frac{1-|z|}{1+|z|},\frac{1+|z|}{1-|z|}\right)$, so $f(z)=g^{-1}(g(f(z))$ lies in that disk. Obviously, the disk lies in $\frac{1-|z|}{1+|z|}\le|w|\le \frac{1+|z|}{1-|z|}$.