Actually, I am trying to solve a mathematical problem, which involves proving an inequality. For that I already know the bounds of LHS of the inequality and now I need the lower bound of the RHS of the inequality (which I mentioned below in the list).
I want a function $F(x)$ such that $F(x) \leq G(x)$, where $G(x)$ is:
- $\sqrt{x} \ln(x) \ln(x)$
- $\frac{1}{8}\pi \sqrt{x} \ln(x) \ln(x)$
- $2 \sqrt{x} \ln(x) \ln(x)$
Here all the above three values are equivalent (in the context of the question) so all or any value that is suitable can be considered for the calculation of lower bound.
But the additional constraint or condition is that $F(x) > H(x)$,
where $H(x) = 0.006409\frac{x}{\ln(x)}$
Note - Kindly note that this is NOT a homework question. I have been trying very hard to find it, so any help will be greatly appreciated. Thanks in advance!
There is no function that satisfies both the inequality for all the value of $\mathbb{R}^+$ as, for big $x$ you will need $$ O\left(\frac{x}{\log(x)}\right) <F(x) < O \left(\sqrt{x} (\log(x))^2\right) $$ that is absurd. Similarly for $x= 1+\epsilon$ you will need
$$ \frac{1+\epsilon}{\log(1+\epsilon)}< F(1+\epsilon)< (\sqrt{1+\epsilon} (\log(1+\epsilon))^2) $$ Taking the limit $\epsilon \to 0^+$ you will reach an absurd.