how to solve the following inequality? I have the following inequality: 
$$n \ge \frac{K_n^2}{\epsilon^2} \frac{\log K_n}{\epsilon},\text{ where }K_n = (\log n)^3.$$
I would like to solve it, even numerically.
I thought that numerically it can be solved by iteratively setting $K_n$ using a current value for $n$ and then $n$ using a current value for $K_n$.
However, I get a weird behavior. For different $\epsilon$ (more specifically for $\epsilon = 3.3409202$) I get that the answer for $n$ is 2.0127e+06. If I increase $\epsilon$ by a bit, I get a complex number. For other similar inequalities, I would get a different, but similar interesting "boundary" behavior: there would be a certain epsilon under which the value of solved $n$ will be very large, and over which the $n$ will suddenly jump to a really small value.
Is there any explanation for what's going on here?
Is there a better way to solve this inequality?
Thanks.

I assume that solving the inequality means replacing the inequality sign by an equality sign and solving for $n$.
From your post, I understand that you run into problems for "large" $\epsilon$ ($\epsilon >3.6$). The equation written explicitly reads
$$\epsilon^3 n = 3 \log^6 n \,\log \log n.$$
It turns out that the equation has two solutions for $\epsilon< \epsilon^*$: one in the region $(e,n^*)$ and the other in the interval $(n^*,\infty)$. For $\epsilon=\epsilon^*$, the only solution is at $n^*$ and for $\epsilon> \epsilon^*$ there is no (real) solution. The values for $\epsilon^*$ and $n^*$ and some explanation you find below:
Let us assume $n\geq 1$ as the right hand side for $n \leq 1$ is in general complex (and also $\epsilon>0$). The right hand side is negative in the region $[1,e]$. As the left hand side is positive, the solution to the equation has to be for $n\geq e$.
The right hand side grows slower than the left hand side (for $n \to \infty$). Furthermore, the left hand side at $n=e$ is $\epsilon^3 e$ whereas the right hand side is 0. Therefore, it is clear that there will be a maximal $\epsilon= \epsilon^*$ for which the equation has a solution (and for $\epsilon < \epsilon^*$ there will be two solutions).
The condition for $\epsilon^*$ is that the left hand side is a tangent (at $n^*$) to the right hand side. In formulas
$$
\begin{align}
\text{LHS}(n^*) &= \text{RHS}(n^*)\\
\text{LHS}'(n^*) &= \text{RHS}'(n^*)
\end{align}
$$
with $\text{LHS}(n)= \epsilon^* n$ and $\text{RHS}(n) = 3 \log^6 n \,\log \log n$. The solution is
$$
\begin{align}
\epsilon^* &= 8.60323 
&n^* &= 687.328 
\end{align}.
$$
