<s>[INST] Prove $\dfrac{n^2}{\ln n^2-1} - \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$ Prove $\dfrac{n^2}{\ln n^2-1} -  \dfrac{\dfrac{n^2}{2}}{\ln \dfrac{n^2}{2}-1.1} > n$ for $n \ge 347$. It seems to be true for all $n \ge 11$, but I only need it to be true for $n \ge 347$. I tried to manipulate the equation in all sorts of ways but I couldn't get it to work. These are the difference of prime bounds if that helps. $\dfrac{x}{\ln x-1} < \pi(x)$ for $x \ge 5393$ and $\pi(x) < \dfrac{x}{\ln x-1.1}$ for $x \ge 60184$ where $\pi(x)$ is counts the number of primes less than or equal to $x$.
 [/INST] I tried to manipulate the equation in all sorts of ways but I couldn't get it to work
In fact, there is an exact solution to the equation
$$\dfrac{n^2}{\log( n^2)-1} - \dfrac{\dfrac{n^2}{2}}{\log \left(\frac{n^2}{2}\right)-a} = n$$ which can rewrite
$$n=4\frac{\left(\log (n)-\frac{1}{2}\right) (\log
   (n)-\alpha )}{\log (n)-\beta }$$ where
$$\alpha=\frac{1}{2} (a+\log (2)) \qquad \beta=\frac{1}{2} (2 a+2 \log (2)-1)$$
Let $n=e^x$ and write
$$e^{-x}=\frac 14 \frac{x-\beta }{\left(x-\frac{1}{2}\right) (x-\alpha )}$$ which shows an exact explicit solution in terms  of the generalized Lambert function (have a look at equation $(4)$).
Since we just look for an estimate, using series
$$e^x=4 x+(2 a-4+2\log
   (2))+O\left(\frac{1}{x}\right) $$ $$x= -\left(\frac{a}{2}+\frac{\log (2)}{2}-1\right)-W_{-1}\left(-\frac{e^{1-\frac{a}{2}}}{4 \sqrt{2}}\right)$$ giving, for $a=1.1$,  $x_0=2.05486$ corresponding to $n_0=7.80572$
Newton iterates are
$$\left(
\begin{array}{cc}
k & x_k & n_k\\
 0 & 2.05486 & 7.80572 \\
 1 & 2.33134 & 10.2917 \\
 2 & 2.30879 & 10.0623 \\
 3 & 2.30852 & 10.0595 \\
\end{array}
\right)$$
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