Let $$f(z) = z - \sum^\infty_{n=2} a_nz^n.$$ What is the largest ball around $0$ where $f$ is injective?
If we restrict to the case where $a_n \geq 0,$ it seems the radius should be given exactly by the minimum positive zero of $f'(x).$ Is there an easy complex analysis proof of this fact in this special case?
Here is a supporting evidence for the conjecture made in the end. Let $f$ be a plynomial (or an entire function). We have $f'=1-P$; where $P$ is a power series with positive coefficients. I claim that the zero of $f'$ which is closest to the origin is positive. Indeed this zero is the closest singularity to the origin of the function $$1/f'=1/(1-P)=1+P+P^2+P^3+... .$$ But this is a power series with positive coefficients, therefore the closest singularity to the origin must be on the positive ray by Pringsheim's theorem.
However, this does not prove the conjecture, even for polynomials, because there can be other obstacles to injectivity, different from a zero of $f'$. One has to prove that these other obstacles cannot occur for such series.
EDIT: In fact the conjecture is true! Let $f(z)=z-Q(z)$, $Q$ has positive coefficients and double zero at zero. Suppose without loss of generality that the zero of $f'$ of the smallest modulus is outside the disc $D$ of radius $r$ with center at the origin.
We wish to prove that $f$ is injective in this disc. Consider $P=Q'$. Then $P(0)=0$ and $P$ has positive coefficients. So $P$ is increasing on $(0,r)$ and cannot take the value $1$. Thus $|P(z)|$ is less than $k$ in $D$, with some $k\in(0,1)$, because coefficients are positive.
Now we prove the statement by contradiction. Suppose $f(z_1)=f(z_2)$ with $z_j$ in the unit disc. Integrating along the straight line segment connecting $z_1$ and $z_2$, we obtain $$0=f(z_2)-f(z_1)=\int_{z_1}^{z_2}f'(\zeta)d\zeta=z_2-z_1-\int_{z_1}^{z_2}P(\zeta)d\zeta.$$ Estimating the integral, we obtain $$|z_1-z_2|=\left|\int_{z_1}^{z_2}P(\zeta)d\zeta\right|\leq k|z_1-z_2|,$$ which is a contradiction. This proves your conjecture.
de Branges's theorem / Bieberbach's conjecture says that if $f$ is injective on the unit disc then $|a_n| \leq n$. Then if $f$ is injective on the ball of radius $r$ than by rescaling $|a_n| \leq n /r^{n_1}$, so we have the upper bound
$r \leq \sup_{n\geq 2} \left( \frac{a_n}{n}\right)^{\frac{-1}{n-1} } $
For all $a_n$ nonnegative, your bound is at least as strong as this one, so this cannot be used to construct counterexamples.
The question can be answered in a general way by http://arxiv.org/abs/1303.6011 is the following way.
You take a function on some polydisk $$f(z) = id(z) - \sum_{|I|\geq2} a_Iz^I.$$ Where $a_I\geq 0.$ As in the paper we're going to evaluate on the commuting tuples of matrices. (This satisfies the conditions on the domains used in the paper) The derivative of f (on commuting tuples of matrices) satisfies the following inequality $$\|Df_i(z)[h]\|\geq Df_i(\|z\|)[\|h\|]$$ (where we're abusing $\|x\| = (\|x_1\|,\ldots,\|x_n\|).)$ Thus, if the derivative is singular, which is equivalent to noninjectivity by the theorem, at some matrix tuple $z$ in some direction $h$, we have that $$0\geq Df_i(\|z\|)[\|h\|]$$ Thus, $$\langle Df(\|z\|)[\|h\|],\|h\| \rangle \leq 0.$$ So $Df[\|z\|]$ has a nonpositive eigenvalue. Thus, on the line segment between $0$ and $\|z\|$ there is a point $w$ such that $Df[w]$ is singular and positive semidefinite. (The derivative at zero is the identity and derivative is real symmetric by assumption) Thus $f$ is not injective on the domain.
