Let $f \colon \mathbb C \to \mathbb C$ be a complex-valued analytic function with non-negative coefficients of Taylor series at 0 (suppose that radius of convergence is $+\infty$ for simplicity): $$ f(z) = c_0 + c_1 z + c_2 z^2 + \ldots \enspace , $$ $c_k \geq 0$ for $k = 0, 1, 2, \ldots$, and also $c_1 > 0$.
It is well-known that for such functions their characteristic function $\phi_0(z) = \dfrac{z f'(z)}{f(z)}$ is nondecreasing when $z \in \mathbb R_+$. We also introduce characteristic function of the derivative $\phi_1(z) = \dfrac{z f''(z)}{f'(z)}$ which is also nondecreasing when $z \in \mathbb R_+$. So, there exists unique real positive solution $\widehat z$ of the equation $\phi_1(\widehat z) = 1$, since $c_1 > 0$ and $\phi_1(0) = 0$, under some mild conditions (there should be at least one nonzero coefficient among $c_3, c_4, \ldots$).
Conjecture. If $0 < |z| < \widehat z$, then it holds $$ |\phi_0(z)| + |2 - \phi_0(z)| \leq 2 \dfrac{f(|z|)}{|f(z)|} \enspace . $$ Moreover, if the function $f(z)$ is aperiodic in the sense $$ \mathrm{GCD}\{i - j \colon c_i \neq 0, \, c_j \neq 0\} = 1 $$ then equality holds only iff $z$ is a real positive, i.e. $z = |z|$.
Hints. I should mention some of my previous attempts on this conjecture. It is important to mention that if $z = |z| < \widehat z$ then $\phi_0(z) \leq 2$. The equality is attained only for some degenerate functions $f(z)$. The sketch of the proof is following. The equation $\phi_1(\widehat z) = 1$ is equivalent to $$ c_1 = 3 c_3 z^2 + 8 c_4 z^3 + \ldots + (k-1)(k+1) c_{k+1}z^k + \ldots \enspace , $$ whereas the inequality $\phi_0(z) \leqslant 2$ is equivalent to $$ 2 c_0 + c_1 z \geq c_3 z^3 + 2 c_4 z^4 + \ldots + (k-2) c_k z^k + \ldots \enspace , $$ which is true because all the coefficients are nonnegative, and $(k-1)(k+1) \geq k-2$.
Some obvious tricks with triangle inequalities didn't work in my case, because there is an obvious lower bound $$ 2 \leq |\phi_0(z)| + |2 - \phi_0(z)| \enspace . $$ I also tried to prove equivalent inequality $$ |z f'(z)| + |2 f(z) - z f'(z)| \leq 2 f(z_0) \enspace , $$ which seems easier, because of explicit representation for $2f(z) - zf'(z)$. If we denote $z = z_0 e^{i \theta}$, then the function $$ \Phi(\theta) = |z f'(z)| + |2f(z) - zf'(z)| $$ can have a plot with large number of local maxima and minima. I plot first summand (1), second summand (2) and lower bound $|2f(z)|$ (3). The sum (1+2) is also depicted. We can see that the behavior is "very tight".
Picture: Separate summands of $\Phi(\theta)$.
Every strategy I tried, didn't seem to give hopeful result, though I tried to simplify and consider particular cases like $f(z) = e^z$, $f(z) = \sinh z$. It is complicated to work with absolute value, since it is not analytic. The idea of $|\omega| = \sqrt{\omega \overline \omega}$ also didn't lead me to the solution, however maybe I was not patient enough to finish that, i mean consider square root of infinite sums. I also tried to simulate the inequality in ipython
for many dfferent sequences, and it always turned out valid.
Motivation. This lemma is important in graph enumeration, because the conjecture leads to some saddle-point estimates for asymptotic number of graphs with certain structure. In particular, the generating function of graph with $n$ unrooted trees, after some transformation, takes the form $g(z)e^{nh(z)}$, and this is the contour integral we want to calculate. Usually, $\Re h(z)$ on a circle $z = z_0 e^{i\theta}$ has maximal value at $z = z_0$, i.e. $\theta = 0$.
Any helpful ideas are appreciated!