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 $x$.

$$\frac{n^2}{\ln n^2-1}-\frac{\frac{n^2}{2}}{\ln\frac{n^2}{2}-1.1} \gt n \; \iff \; 
\frac{1}{\ln n^2-1}-\frac{1}{2(\ln\frac{n^2}{2}-1.1)} \gt \frac{1}{n} \tag{1}\label{eq1A}$$
For $n \ge 347$, we have $\ln(n) \gt 2(\ln(2) + 1.1) \; \to \; \ln(n) - \ln(2) - 1.1 \gt \frac{1}{3}\left(2\ln(n) - \ln(2) - 1.1\right)$ so, using this below, we get
$$\begin{equation}\begin{aligned}
\frac{1}{\ln n^2-1}-\frac{1}{2(\ln\frac{n^2}{2}-1.1)} & = \frac{1}{2\ln(n)-1}-\frac{1}{2(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{2(2\ln(n) - \ln(2)-1.1) - (2\ln(n)-1)}{(2\ln(n)-1)2(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{2\ln(n) - 2\ln(2)- 1.2}{(2\ln(n)-1)2(2\ln(n) - \ln(2)-1.1)} \\
& \gt \frac{2\ln(n) - 2\ln(2)- 2(1.1)}{(2\ln(n)-1)2(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{\ln(n) - \ln(2)- 1.1}{(2\ln(n)-1)(2\ln(n) - \ln(2)-1.1)} \\
& \gt \frac{2\ln(n) - \ln(2)- 1.1}{3(2\ln(n)-1)(2\ln(n) - \ln(2)-1.1)} \\
& = \frac{1}{3(2\ln(n)-1)}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note that
$$\frac{1}{3(2\ln(n)-1)} \gt \frac{1}{n} \iff 3(2\ln(n)-1) \lt n \iff n - 6\ln(n) + 3 \gt 0 \tag{3}\label{eq3A}$$
Thus, to prove \eqref{eq1A}, it's sufficient to prove the RHS above. Define
$$f(n) = n - 6\ln(n) + 3 \tag{4}\label{eq4A}$$
We have $f(347) \approx 314.9$, and $f'(n) = 1 - \frac{6}{n} \gt 0$, so $f(n) \gt 0$ for all $n \ge 347$.
