I was playing around with the prime counting function and came across something that seemed correct to me, maybe it's already been proven but I don't know so I decided to ask here.
maybe a stupid question but whatever
Is $$\pi((n+1)^2) < \frac{(n+2)^2}{\ln((n+2)^2)}$$ for $n>0$?
A numerical calculation (in, say, Wolfram Alpha or Mathematica) shows that if $x \geq e$, then $$\frac{(\sqrt{x}+1)^2}{2\log(1+\sqrt{x})}<\frac{x}{\log x}\Big(1+\frac{1}{\log x}+\frac{2}{(\log x)^2}\Big).$$ Moreover, Pierre Dusart proved that if $x\geq 88,879$, then $$ \frac{x}{\log x}\Big(1+\frac{1}{\log x}+\frac{2}{(\log x)^2}\Big)<\pi(x). $$ Therefore, if $x\geq 88,879$, then $$ \frac{(\sqrt{x}+1)^2}{2\log(1+\sqrt{x})}<\pi(x). $$ Under the change of variables $n=\sqrt{x}-1$, we find that if $n\geq 297$, then $$ \frac{(n+2)^2}{\log[(n+2)^2]}<\pi((n+1)^2). $$ ADDED: An additional numerical calculation shows that the only positive integers $n$ such that $$\pi((n+1)^2)<\frac{(n+2)^2}{\log[(n+2)^2]}$$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, and $9$.
