Proof that $\dfrac{n(\ln(n)-2\ln(2))}{2\ln(n)(\ln(n)-\ln(2))^2} > 1$ for all $n\ge20$ How to prove this inequality: $\dfrac{n(\ln(n)-2\ln(2))}{2\ln(n)(\ln(n)-\ln(2))^2} > 1$ for all $n\ge20$
I tried to apply this approach but I get a large first differentiate $u'(x)$ whose sign is not easy to determine
 A: Idea:  The goal is to show that $f(x)$ is increasing.  We could take the derivative, but in this case that will be tedious.  Instead use logarithms.  We have that $f(x)$ is positive and increasing if and only if $\log f(x)$ is increasing, so all we need to do is look at the logarithmic derivative and the value at $n=20$.
Proving the Inequality: Rewriting your above fraction, let $$g(n)=\frac{n\log\left(\frac{n}{4}\right)}{\log\left(n\right)\left(\log\left(\frac{n}{2}\right)\right)^{2}}.$$ then $$\log\left(g(n)\right)=\log n+\log\log\left(\frac{n}{4}\right)-\log2-\log\log n-2\log\log\left(\frac{n}{2}\right).$$ Taking the derivative we see that $$\frac{g^{'}(x)}{g(x)}=\frac{1}{x}+\frac{1}{x\log\left(\frac{x}{4}\right)}+\frac{-1}{x\log x}+\frac{-2}{x\log\left(\frac{x}{2}\right)}.$$ Now, notice that $\frac{1}{\log\left(\frac{x}{4}\right)}+\frac{-1}{\log x}>0$. Next, for $x\geq20$,  $\log\left(\frac{x}{2}\right)\geq\log\left(10\right)>\log(e^{2})=2$  so that $1+\frac{-2}{\log\left(\frac{x}{2}\right)}>0.$ This implies that for $x\geq20$ $$\frac{g^{'}(x)}{g(x)}>0.$$  As $g(20)>1$ , the desired inequality follows.
