Show $|f(z)| \leq \frac{2A|z|}{1 - |z|}$ I have the following question:

Let $f$ be a holomorphic function on $\mathbb{D}$. Assume $f(0) = 0$ and $\Re f ≤ A$ on $\mathbb{D}$ for
some constant $A > 0$. Show that
$|f(z)| ≤ \frac{2A|z|}
{1 − |z|}$
for all $z ∈ \mathbb{D}$.

The $f(0)$ hypothesis and the fact that $f$ is defined on $\mathbb{D}$ makes me think that this could involve some type of Schwartz Lemma. But, the there is no reason to think that the function is bounded by $1$. I also do not know how to dissect the $\Re f \leq A$ part.
Any help would be greatly appreciated.
 A: Wlog we can assume that $f$ extends to the boundary analytically (as otherwise we use $f_r(z)=f(rz)$ and take $r \to 1$) and we use what Wikipedia calls the Schwarz Integral formula (or what is called in some books the Poisson-Cauchy representation - the Cauchy completion of the Poisson formula):
$f(z)=\frac{1}{2\pi}\int_0^{2\pi}{\Re f(e^{it})\frac{e^{it}+z}{e^{it}-z}}dt$ so
$f(z)=f(z)-f(0)=\frac{1}{2\pi}\int_0^{2\pi}{\Re f(e^{it})(\frac{e^{it}+z}{e^{it}-z}}-1)dt=\frac{1}{2\pi}\int_0^{2\pi}{\Re f(e^{it})\frac{2z}{e^{it}-z}}dt$
Taking absolute values:
$|f(z)| \le \frac{1}{2\pi}\int_0^{2\pi}{A\frac{2|z|}{|e^{it}-z|}}dt$
But $|e^{it}-z| \ge 1-|z|$ so we get the required result and we are done:
$|f(z)| \le \frac{1}{2\pi}\int_0^{2\pi}{A\frac{2|z|}{1-|z|}}dt=A\frac{2|z|}{1-|z|}$
A second solution can be given noting that $\Re (-f/A) \ge -1, f(0)=0$ means that $f$ is subordinated to $g(z)=\frac{2z}{1-z}$ (which sends the unit disc precisely to the domain $ \Re w >-1$ as wlog we can assume $Re f <A$ as otherwise $f=0$ by maximum modulus and $f(0)=0$)
This means that there is $\phi(0)=0, \phi : D \to D$ holomorphic so $-f(z)/A=g(\phi(z))=\frac{2\phi(z)}{1-\phi(z)}$.
But $|\phi(z)| \le |z|$ by Schwarz so taking absolute values we recover the required result since $|1-\phi(z)| \ge 1 -|\phi(z)| \ge 1-|z|$
