Are there numbers $k > 1$ and $c > 1$ such that:


1 ) $\theta(c) \geq c \left( 1-\frac{1}{5 \ln^2(c)} \right) $


2 ) $\frac{c}{1+\frac{1}{\ln^4(c)}} \leq p(\pi(c))$ where $p(n)$ is the $n$-th prime number and $\pi(n)$ is the prime counting function


3 ) $\pi(c) \leq \frac{c}{\ln (c)} (1+\frac{1}{\ln (c)}+\frac{2}{\ln^2 (c)}+\frac{6}{\ln^3 (c)} + \frac{25}{\ln^4 (c)} )$


4 ) $\pi(c) \geq \frac{c}{\ln (c)} (1+\frac{1}{\ln (c)}+\frac{2}{\ln^2 (c)}+\frac{6}{\ln^3 (c)} + \frac{24}{\ln^4 (c)} )$


5 ) $\prod \limits\_ {p \leq c} (1+\frac{1}{p^2-1}) \prod \limits\_{p \leq c^k} (1+\frac{1}{p}) \leq e^\gamma \ln (c ( 1-\frac{1}{5 \ln^2(c)})+c^k ( 1-\frac{1}{5 k^2 \ln^2(c)}))$?



Numbers related to the Riemann hypothesis
Proving such results falls into three parts. First you take an effective version of the prime number theorem, which implies all your desired bound for sufficient. Second you write a computer program (or use computations done by somebody else) to check your inequalities for small $x$. Finaly you bridge the gap between the two ranges by improving the general bounds for large $x$ with more specialized arguments using numerical information on zeros of the Riemann $\zeta$-function.


The first part is easy, since you can just cite the error terms of the effective PNT. As far as I know, the bet results are due to Trudgian ( <https://arxiv.org/pdf/1401.2689.pdf> ). The difficult part is making the different ranges overlap. Note that as your inequalities are rather weak, and $\pi(x)$ can be computed in sublinear time, you should compute $\pi(x\_0)$, use an upper bound sieve to compute some $x\_1$, such that your inequality holds in $[x\_0, x\_1]$. Since $x\_1$ will be of magnitude $x\_0+\frac{x\_0}{\log^C x\_0}$, the savings are substantial.


