Is it true that $$\prod_{p\le x}\frac p{p1}\le e^\gamma\ln x\left(1\frac{0{.}011}{\ln x}+\frac{0.2}{(\ln x)^2}\right)$$ for all $x>25\,000$, where the product is over prime $p$?

1$\begingroup$ where is the question? $\endgroup$– kodluSep 6, 2016 at 8:26

$\begingroup$ is the inequality holds for all integers bigger than 25000 $\endgroup$– AhmadSep 6, 2016 at 8:44

1$\begingroup$ I took the liberty to reformat the post so that it reads like an actual question, in English. I hope I got the meaning correctly. $\endgroup$– Emil JeřábekSep 6, 2016 at 9:08
1 Answer
Your inequality fails for every $x\geq e^{700/11}\approx 4.33433\times 10^{27}$.
Indeed, by Theorem 8 in the classic paper RosserSchoenfeld: Approximate formulas for some functions of prime numbers, we have $$e^\gamma\ln x\left(1\frac{0.5}{(\ln x)^2}\right)<\ \prod_{p\le x}\frac p{p1},\qquad x>1,$$ hence your inequality can only hold when $$1\frac{0.5}{(\ln x)^2}< 1\frac{0{.}011}{\ln x}+\frac{0.2}{(\ln x)^2},$$ i.e., when $$\ln x< 0.7/0.011=700/11.$$
Added. In fact Diamond and Pintz showed (Journal de Théorie des Nombres de Bordeaux 21 (2009), 523533) that the product in question exceeds $e^\gamma\ln x$ for infinitely many integers $x$ (actually they prove much more than this).