Is the function $\frac{\operatorname{Harmonic}(x)}{\operatorname{Airy}(x)}$ a convex function for real numbers $0<x<1$?

Question

For real numbers $0\leq x\leq 1$, I know how to exlore some aspects of the graph ( I was playing with Wolfram Alpha online calculator about its plot, see next paragraph) of the function $$f(x)=\frac{H_x}{\operatorname{Ai}(x)},$$ where $\operatorname{Ai}(x)$ is the Airy function, see this MathWorld and $H_x$ is the harmonic number.

The code in Wolfram Language is

plot Harmonic(x)/Airy(x), for 0<x<1

But I don't know if it is possible to prove that this function is convex $f''(x)\geq 0$ for $0<x<1$. I've calculated the second derivative with Wolfram Alpha online calculator with this code

d^2/dx^2 Harmonic(x)/Airy(x)

Question. Is it possible to prove that $$\frac{H_x}{\operatorname{Ai}(x)}\tag{1}$$ is a convex function for real numbers in the open interval $(0,1)$? Many thanks.

Answer 1

Mathematica gives $$f''(0)=\frac{\pi \left(\pi \Gamma \left(\frac{2}{3}\right)^2-4 \sqrt[6]{3} \zeta (3)\right)}{\Gamma \left(\frac{1}{3}\right)} \approx -0.0162334,$$ and continuity implies that the second derivative remains negative near $x=0$. Thus $f$ is not convex over the interval $(0,1)$.

Answer 2

I would say it was a close call: Supporting user1337's computation, here is a graph of the second derivative on $[0, 0.002]$ is as follows:

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