Show that $D_{\alpha/M}(0)$ is contained in the image I've been trying to do this problem for five-ever but haven't gotten anywhere.

Show there exists an $\alpha > 0$ such that for any $h \in \mathcal O(D)$ ($D$ being the unit disc) with $M = \sup\limits_{z \in D} |h(z)| < \infty$, $h(0) = 0$, and $h'(0) = 1$, the ball $D_{\alpha/M}(0)$ is contained in the image $h(D)$.

All I've got is that since $|h(z)| < M$, the function $\frac{h(z)}{M}$ maps $D$ into itself, so $|h(z)|/M \leq |z|$ by the Schwarz lemma.  Also, $h'(0)/M < 1$, so necessarily $M > 1$.  
I also know that $h$ is locally injective near $0$.
 A: If $\lvert h(z) - z\rvert \leqslant \frac{\lvert z\rvert}{2}$ for $\lvert z\rvert \leqslant r$ (where $0 < r < 1$), then we have
$$D_{r/2}(0) \subset h\bigl(D_r(0)\bigr)$$
by the argument principle. For we then have $\lvert h(z)\rvert \geqslant r/2$ on $\{ z : \lvert z\rvert = r\}$, and hence
$$N(w) = \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{h'(z)}{h(z)-w}\,dz$$
is constant on $D_{r/2}(0)$. Since $N(0) \geqslant 1$, the assertion follows.
So let's see how far we can go while being sure that $\lvert h(z) - z\rvert \leqslant \frac{\lvert z\rvert}{2}$. We have
$$h(z) - z = z\cdot\int_0^1 h'(tz) - 1\,dt.\tag{1}$$
Now $$h'(w) - 1 = w\cdot\int_0^1 h''(uw)\,du,\tag{2}$$ and by the integral formula
$$h''(\zeta) = \frac{2}{2\pi i} \int_{\lvert z\rvert = \rho} \frac{h(z)}{(z-\zeta)^3}\,dz$$
for $\lvert\zeta\rvert < \rho$, we obtain the estimate
$$\lvert h''(\zeta)\rvert \leqslant \frac{2}{2\pi}\frac{2\pi \rho}{(\rho - \lvert\zeta\rvert)^3} \cdot M.\tag{3}$$
Since $(3)$ holds for all $\lvert \zeta\rvert < \rho < 1$, we obtain
$$\lvert h''(\zeta)\rvert \leqslant \frac{2M}{(1-\lvert \zeta\rvert)^3}.\tag{4}$$
Inserting $(4)$ into $(2)$ we obtain
\begin{align}
\lvert h'(w) - 1\rvert &\leqslant \lvert w\rvert\cdot \int_0^1 \frac{2M}{(1-t\lvert w\rvert)^3}\,dt\\
&= 2M\int_0^{\lvert w\rvert} \frac{du}{(1-u)^3}\\
&= \biggl[\frac{M}{(1-u)^2}\biggr]_0^{\lvert w\rvert}\\
&= M\biggl(\frac{1}{(1-\lvert w\rvert)^2} - 1\biggr)\\
&\leqslant \frac{2M\lvert w\rvert}{(1-\lvert w\rvert)^2}.
\end{align}
Using this estimate in $(1)$, we obtain
\begin{align}
\int_0^1 \lvert h'(tz) - 1\rvert\,dt &\leqslant \int_0^1 \frac{2M\lvert tz\rvert}{(1-\lvert tz\rvert)^2}\,dt\\
&= \frac{2M}{\lvert z\rvert} \int_0^{\lvert z\rvert} \frac{u}{(1-u)^2}\,du\\
&= \frac{M}{\lvert z\rvert} \Biggl(\biggl[\frac{u^2}{(1-u)^2}\biggr]_0^{\lvert z\rvert} - \int_0^{\lvert z\rvert} \frac{2u^2}{(1-u)^3}\,du\Biggr)\\
&\leqslant \frac{M\lvert z\rvert}{(1-\lvert z\rvert)^2}.
\end{align}
From that we find that certainly
$$\int_0^1 \lvert h'(tz)-1\rvert\,dt \leqslant \frac{1}{2}$$
for $\lvert z\rvert \leqslant \frac{1}{2(M+1)}$. Since $M \geqslant 1$ by the Schwarz lemma, we thus can choose $\alpha = \frac{1}{8}$.
