Numbers related to the Riemann hypothesis

Question

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

Answer 1

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.