The following is an old qualifying exam question. The unit disk $\{z \in \mathbb{C} : |z| < 1\}$ is denoted by $\mathbb{D}$.
For an analytic function $f:\mathbb{D}\to\mathbb{D}$, if $f(0) = \alpha$ and $f'(0) = 0$, then for $|z| \le \sqrt{|\alpha|}$, $$|f(z)| \ge \frac{|\alpha| - |z|^2}{1 + |\alpha||z|^2}$$
My hunch is to use the Cauchy integral formula -- for $C$ the counter-clockwise closed circular curve centered at 0 of radius $\sqrt{|\alpha|}$ and fixed $w$ inside $C$, $$ f(w) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - w} dz$$ which gives us $$f(0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z} dz = \alpha \qquad \text{ and } \quad f'(0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z^2} dz = 0$$ From here, I'm not seeing anything immediate. Any ideas?
$$g(z) = \frac{f(z)-\alpha}{1- \overline \alpha f(z)}$$ maps $\Bbb D$ into itself, with $g(0) = g'(0) = 0$.
Applying the Schwarz Lemma twice (to $g(z)$ and $g(z)/z$) it follows that $$ \left\vert \frac{f(z)-\alpha}{1- \overline \alpha f(z)} \right \vert \le |z|^2 \, . $$ Therefore $$ |\alpha| - |f(z)| \le |z|^2 ( 1 + |\alpha| |f(z)|) \, , $$ from which the desired conclusion follows.