I am trying to solve a graph problem where $n \geq 1$. I have found an online article here, which uses: $$\ln 6 + \ln \ln n - \ln \ln \ln n - 1 \geq \frac{\ln \ln n}{2}$$
I find it difficult to understand this part. I am rewriting the left hand side as: $$ \ln \left( \frac{6\ln(n)}{\mathbf{e}\ln(\ln(n))} \right) $$ and now the inequality is $$ \ln \left( \frac{6\ln(n)}{\mathbf{e}\ln(\ln(n))} \right) \geq \frac{\ln \ln n}{2} $$ If I take the exponentials of both sides, it can help but then things become even more complicated, since if left side is negative, then taking exponentials will change the inequality etc...
Extra: In the same link, there is also a very similar step, which says for $\alpha = \tfrac{6\ln(n)}{\ln\ln(n)}$ we can show $-\alpha\ln(\alpha) + \alpha - 1 < 0$, and I think even $-\alpha\ln(\alpha) + \alpha $ is negative. I am trying to take the derivatives, hessians etc to see the behavior, but these are really dependent on $n$ and it seems that $n \geq \mathbf{e}$ is needed. Even with that, I could not prove... I thought I need some help.
Suppose $n=e$. Then, the LHS is undefined (division by zero). So, you must have $n>e$. Anyway, multiply both sides by 2 to get:
$$\ln\left(\dfrac{6\ln(n)}{e\ln(\ln(n))}\right)^2 \ge \ln\ln n$$
Take out the outer-most $\ln$ to get
$$\left(\dfrac{6\ln(n)}{e\ln(\ln(n))}\right)^2 \ge \ln n$$ Divide both sides by $\ln n$:
$$\dfrac{36\ln(n)}{e^2\left(\ln(\ln(n))\right)^2} \ge 1$$
That is a bit easier to prove.
$$36\ln n - e^2\left(\ln (\ln n)\right)^2 \ge 0$$
Let $f(n) = 36\ln n - e^2\left(\ln(\ln n)\right)^2$.
Taking the derivative with respect to $n$ and simplifying to find when the derivative is zero gives:
$$36\ln n - 2e^2\ln(\ln n) = 0$$
$$n^{36} = \left(\ln n\right)^{2e^2}$$
However, we know:
$$n^{36} \ge (\ln n)^{36} \ge (\ln n)^{2e^2}$$
So, it will never be zero. Plugging in $e^e$ gives a positive value for the derivative, so the derivative is always positive and $f(n)$ is increasing.
$n=e^e$, we have:
$$f(e^e) = 36e - e^2>0$$
Therefore, $f(n)\ge 0$ for all $n>e$.