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 $\Phi(\theta)$.](https://i.stack.imgur.com/roXqS.png)


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!



Analytic Combinatorics: upper bound for sum of absolute values of two complex functions: $|z f'(z)| + |2 f(z) - zf'(z)| \leq 2f(|z|)$
I think, it is true and follows from the following elementary


**Lemma.** If $\alpha>\beta>0$ and $1/\beta-1/\alpha\geqslant 2$, then $|1+\alpha z|+|1-\beta z|\leqslant 2+\alpha-\beta$ for any vector $z$ with $|z|=1$.


**Proof** of the lemma. Some nice argument must exist, but you may simply denote $z=e^{it}$ and differentiate LHS in $t$. For real $z$ the inequality is clear, else in a point where derivative equals 0 you get $\alpha/|1+\alpha z|=\beta/|1-\beta z|$, $|z+1/\alpha|=|z-1/\beta|$, but perpendicular bisector to the segment $[-1/\alpha,1/\beta]$ does not contain non-real points of a unit circle provided that $1/\beta-1/\alpha\geqslant 2$.


Now we have $c\_1\geqslant \sum\_{k\geqslant 2} (k^2-1)c\_{k+1}|z|^k$ and need to prove 
$$
|c\_1z+2c\_2z^2+3c\_3z^3+\dots |+|2c\_0+c\_1z-c\_3z^3-2c\_4z^4-\dots|\leqslant\\ 2c\_0+2c\_1|z|+2c\_2|z|^2+\dots,
$$
for proving that at first we use triangle inequality for removing terms with $c\_0$ and $c\_2$, then divide by $|z|$, we reduce our inequality to
$$
|c\_1+3c\_3z^2+\dots |+|c\_1-c\_3z^2-2c\_4z^3-\dots|\leqslant 2c\_1+2c\_3|z|^2+\dots,
$$
set $c\_1=a+3c\_3|z|^2+8c\_4|z|^3+\dots$ for some non-negative $a$ and use triangle inequality many times, we reduce the inequality to a family of inequalities $$|(k^2-1)|z|^k+(k+1)z^k|+|(k^2-1)|z|^k-(k-1)z^k|\leqslant (2(k^2-1)+2)|z|^k$$
(this inequality goes with coefficient $c\_{k+1}$), but this is a partial case of lemma with $\alpha=1/(k-1),\beta=1/(k+1)$.


