I'm having trouble proving the following:
Let $f$ be analytic on $D(0,R)$ and suppose there is a real constant A for which $\operatorname{Re}f(z) < A$ for all $z \in D(0,R)$. If $f(0) = 0$, show that $|f(z)| \leq \frac{2A|z|}{R-|z|}$.
I was told to first start by showing that $|\frac{f(z)}{2A-f(z)}| \leq 1$ and also use the following result, which I was able to prove:
Let $f$ be analytic on $D(0,R)$ and let $f(0)=0$. If $|f(z)| \leq M$ for all $z \in D(0,R)$, then $|f(z)| \leq M|z|/R$ for all $z\in D(0,R)$.
Thanks in advance!
Apply the form of Schwarz Lemma you have already proved to the function $\frac {f(z)} {2A-f(z)}$ (with $M=1$). You will get $|\frac {f(z)} {2A-f(z)}| \leq \frac {|z|} R$. Cross multiply and use the inequality $|2A-f(z)| \leq 2A+|f(z)|$. (To see that the function is bounded by $M=1$ you have to show that $|\frac {a+ib} { 2A-(a+ib)}| \leq 1$ if $a$ and $b$ are real and $a<A$. This is easy since $|2A-(a+ib)|^{2} =(2A-a)^{2}+b^{2} \geq a^{2}+b^{2}$ because $|2A-a| \geq |a|$).