Proof that $\oint_r d(x,N + n) < 0 $?

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Let $f(x)$ be a real-entire function such that for all $x>0$ we have $\DeclareMathOperator{\dmu}{d\!}$,

  • $f(x) > 0$, $f'(x) > 0$ , $f '' (x) > 0$,

  • and also $0 < D^M f(0) < D^{M-1} f(0)$,

Furthermore, put $g(x,n) = \frac{f(x)}{x^n}$ where $n$ is a positive integre and for $0<T<1$ let

  • $x_0(n)$ be a real satisfying $x_0 (n)> 0$ and such that $$ g ' (x_0(n),n) = 0, $$
  • $r(n)$ be a contour that contains the real interval $[T,x_0(n) + T]$ but not $0$ , not a negative number and no poles or zero's off the real line.

Conjecture A: there is Always an integer $N$ ( depending on $f$ ) such that for all $n > N$ $$ \frac {1} {2 \pi i} \oint_{r(n)} \bigg[(x-T)^{n-1} f(x+T) - \sqrt n \ln(e+n) g(x,n) \frac{g '' (x,n) }{g ' (x,n) }\bigg]\dmu x < 0. $$

A similar conjecture that is probably not compatibel with Conjecture A (they probably cannot both be true , but I believe at least one is true).

Conjecture B: there is Always an integer $N$ ( depending on $f$ ) such that for all $n > N$ there exists a positive $w$ independent of $n$ such that $$ \frac {1} {2 \pi i} \oint_{r(n)} \bigg[(x-T)^{n-1} f(x+T) - \frac{w g(x,n) g '' (x,n) \sqrt { f '' (x)} }{g ' (x,n) }\bigg]\dmu x < 0. $$

( the derivatives are with respect to $x$ , not $n$ )

How to prove/disprove them ?

In case these contour integrals look random/confusing notice that

$$ \frac {1} {2 \pi i} \oint_{r(n)} (x-T)^{n-1} f(x+T)\dmu x $$

is just the sequence of Taylor coëfficiënts for $f(x)$.
Also $$ \frac {1} {2 \pi i} \oint_{r(n)} g(x,n) \frac{g '' (x,n) }{g ' (x,n) }\dmu x$$

is simply minimum( $f(x)/x^n $) for $x > 1.$
So basically we want to understand how good the intuïtive estimate min( $f(x)/x^n $) is compared to $D^n f(x) / n! $.

EDIT
Maybe this helps ??

https://en.wikipedia.org/wiki/Wiman-Valiron_theory