In Dusart papers he proves that $\prod \limits\_{p \leq x} \frac{p\_i}{p\_i-1} \leq e^\gamma \ln (x) \left(1+\frac{0.2}{\ln ^2 (x)} \right)$ for large numbers.


What I am asking is could we make the bound a little bit sharper by making it like this 
$$ \prod \limits\_{p \leq x} \frac{p\_i}{p\_i-1} \leq e^\gamma \ln (x) \left(1+\frac{0.06}{\ln ^3 (x)} \right)$$ for sufficiently large numbers? 



Even sharper upper bound for prime product?
Yes, although writing down explicitly what "sufficiently large" means might be a challenge.


The prime number theorem implies that there exists a constant $c>0$ such that
$$
\prod\_{p\le x} \frac p{p-1} = e^\gamma \ln (x) \big( 1 + O\big( \exp(-c\sqrt{\ln x}) \big) \big).
$$
From this it is easy to deduce, for every $\varepsilon>0$ and every $A>0$, that
$$
\prod\_{p\le x} \frac p{p-1} \le e^\gamma \ln (x) \bigg( 1 + O\bigg( \frac{\varepsilon}{\ln^A(x)} \bigg) \bigg)
$$
holds when $x$ is sufficiently large in terms of $\varepsilon$ and $A$. (Also, the right-hand side with $+$ changed to $-$ is a lower bound for the prime product for sufficiently large $x$.)


