Let $F:\mathbb{N} \to \mathbb{N}$ be a function that is neither injective nor surjective.
Let $a, b$ be two elements in the domain of $F$.
QUESTION
Suppose that the function $F$ satisfies (at the two elements $a$, $b$) the following inequalities: $$C \leq \frac{1}{2}\cdot\bigg(\frac{F(a)}{b} + \frac{F(b)}{a}\bigg) < \frac{b}{F(a)} + \frac{a}{F(b)}$$ (for some absolute numerical constant $C$) and $$\frac{1}{2} < \frac{ab}{F(a)F(b)} < \frac{1}{\sqrt[3]{2}}.$$
(Note that we can take $C = \sqrt[6]{2}$ by using the Arithmetic Mean-Geometric Mean Inequality.)
Does it follow that $$\frac{b}{F(a)} + \frac{a}{F(b)}$$ is bounded from above?
MY ATTEMPT
I tried to rewrite the second set of inequalities as: $$\frac{1}{2}\cdot\frac{F(b)}{a} < \frac{b}{F(a)} < \frac{1}{\sqrt[3]{2}}\cdot\frac{F(b)}{a}$$ and $$\frac{1}{2}\cdot\frac{F(a)}{b} < \frac{a}{F(b)} < \frac{1}{\sqrt[3]{2}}\cdot\frac{F(a)}{b}$$ but I appear to be going nowhere close to getting an upper bound for $$\frac{b}{F(a)} + \frac{a}{F(b)}.$$
For fixed $a,b$ and an arbitrary $F$ the obvious upper bound is $\displaystyle\frac{b}{F(a)} + \frac{a}{F(b)}\le a+b\,$.
For arbitrary $a,b,F$ there is no upper bound independent of $\,a,b,F$. Consider for example $\,a=u^2, b=4v^2\,$, $\,F(a)=2v\,$, $\,F(b)=3 u^2 v\,$. The conditions of the problem are satisfied:
$\displaystyle \frac{F(a)}{b} + \frac{F(b)}{a} = \frac{1}{2v}+3v \ge \sqrt{6}\,$ by AM-GM, so $\displaystyle\,C = \frac{\sqrt{6}}{2}\,$ works;
$\displaystyle \frac{ab}{F(a)F(b)} = \frac{4}{6} = \frac{2}{3} \in \left(\frac{1}{2}, \frac{1}{\sqrt[3]{2}}\right)\,$.
But $\displaystyle \frac{b}{F(a)} + \frac{a}{F(b)}=2v+\frac{1}{3v}$ can be made arbitrarily large by choosing a large enough $v\,$.