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



Mertens' 3rd theorem, upper bound
Your inequality fails for every $x\geq e^{700/11}\approx 4.33433\times 10^{27}$.


Indeed, by Theorem 8 in the classic paper Rosser-Schoenfeld: Approximate formulas for some functions of prime numbers, we have
$$e^\gamma\ln x\left(1-\frac{0.5}{(\ln x)^2}\right)<\ \prod\_{p\le x}\frac p{p-1},\qquad x>1,$$
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), 523-533) that the product in question exceeds $e^\gamma\ln x$ for infinitely many integers $x$ (actually they prove much more than this).


