Suppose that $f$ is analytic in the unit disk and satisfies Re$\,f(z)>0$ and $\,f(0)=c$. Show that
$$
\frac{1-|z|}{1+|z|}\le \,f(z)|\le\frac{1+|z|}{1-|z|}.
$$
How do we prove this proof this? I know that we should define the function $F(z)=\dfrac{f(z)-1}{f(z)+1}$.
Hint. First of all, in order for the inequality to hold, $|c|$ has to be equal to 1.
Set $$ g(z)=\frac{f(z)-c}{f(z)+c}. $$ Then $\,\mathrm{Re}\,f(z)>0$, implies that $|g(z)|<1$. Also, $g(0)=0$, and by virtue of Schwarz Lemma, $|g(z)|\le |z|$. Thus $$ \left|\frac{f(z)-c}{f(z)+c}\right|\le |z|. $$ Hence $$ \frac{1-|f(z)|}{1+|f(z)|}\le |z|\quad\Longrightarrow\quad \frac{1-|z|}{1+|z|}\le|f(z)| $$ and $$ \frac{|f(z)|-1}{|f(z)|+1}\le |z|\quad\Longrightarrow\quad \frac{1+|z|}{1-|z|}\ge|f(z)|. $$