Solve the inequality for $n\in\mathbb N$: $$\frac{n}{1,785\ln(\ln n)+\frac{3}{\ln(\ln n)}} < 8$$
Attempt:
After some algebra, $$\frac{n\ln(\ln n)-14,28\ln(\ln n)\ln(\ln n)-24}{1,785\ln(\ln n)\ln(\ln n)+3}<0$$
$$n\ln(\ln n)-14,28\ln(\ln n)\ln(\ln n)-24<0;1,785\ln(\ln n)\ln(\ln n)+3\neq 0$$
How to solve this inequality (without the use of computer)?
Note: I have found the inequality in this post
and I am using it to find all positive integers $n$ such that $\phi(n)=8$.
This may not be a satisfactory answer for you.
The term $\log(\log(n))$ varies extremely slowly in the range $10\leq n\leq 50$. I computed its average value over that range and it is "almost" $1.2$. So, for a preliminary search, you could use $$\frac{n}{e^{\gamma } \log (\log (n))+\frac{3}{\log (\log (n))}}\approx \frac{n}{1.2e^{\gamma } +\frac{3}{1.2}}\approx 0.215n$$ which would give $n=37$ while the exact solution would be $n=37$ !!
In fact, you could perform a linear regression and obtain, for this range $$\frac{n}{e^{\gamma } \log (\log (n))+\frac{3}{\log (\log (n))}}\approx 0.220034 n-0.152421$$
