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!