Inequality involving a product over the primes Is there someone who is able to prove the following statement?
$$\prod_{m=1}^n \dfrac{p_m-1}{p_m} \leq \dfrac{1}{\ln(n)}$$ for all integers $n >1$ where $p_m$ is the $m$-th prime number.
 A: We have $$\prod_{i\leq n}\frac{p_{i}-1}{p_{i}}=\prod_{i\leq n}\left(1-\frac{1}{p_{i}}\right)=\frac{1}{\log\left(p_{n}\right)e^{\gamma}}+O\left(\frac{1}{\log^{2}\left(p_{n}\right)}\right)
 $$ by the Mertens theorem. Now note that, by Rosser's theorem $$\frac{1}{\log\left(p_{n}\right)e^{\gamma}}<\frac{1}{\log\left(p_{n}\right)}<\frac{1}{\log\left(n\log\left(n\right)\right)}=\frac{1}{\log\left(n\right)+\log\left(\log\left(n\right)\right)}\leq\frac{1}{\log\left(n\right)}.
 $$
A: Below I've left my previous flawed approach. It is interesting that my upvoters and I hadn't noticed the highlighted mistake. And also that there exists a very straightforward proof: $$\prod_{m=1}^n\frac{p_m}{p_m-1}-\ln n>\sum_{m=1}^n\frac{1}{m}-\ln n>\gamma>0. $$

If you wish to avoid Mertens but not Rosser, note that your inequality is weaker than $$\begin{align}\prod_{m=1}^n\frac{p_m-1}{p_m}&\le\frac{n}{p_n} \\ (p_n-1)\prod_{m=1}^{n-1}\frac{p_m-1}{p_m}&\le n \\ p_n-1&\le n\prod_{m=1}^{n-1}\frac{p_m}{p_m-1}, \end{align}$$which holds for $n=2$ and for larger $n$ follows from  $$\begin{align} \color\red{\frac{p_{n+1}-1}{p_n-1}}\require{enclose}  \enclose{updiagonalstrike,downdiagonalstrike}{\frac{p_n}{p_n-1}} &<\frac{n+1}{n} \frac{p_n}{p_n-1}    \\ \color\red{p_n}\left(1+\frac{1}{n}\right)&>\enclose{updiagonalstrike,downdiagonalstrike}1\color\red{p_{n+1}-1} \\ \color\red{p_{n+1}-p_{n}} &\color\red{<}\color\red{\frac{p_n}{n}+1} \longrightarrow \text{stronger than Cramer's conjecture}\end{align}$$In fact, too stronger. It fails infinitely often, as a consequence of Westzynthius's result (kindly reminded by wythagoras in the comments): $$\lim\sup\frac{p_{n+1}-p_n}{\log p_n}=\infty.$$ 
