Question about bounding a product over primes Let $a$ be a non-negative integer. 
Is it true that $$ \prod_{p\leq x} \left(1+\frac{a}{p}\right)\leq \left(\frac{\log{x}}{\log{2}}\right)^a?$$
Edit: If we apply the logarithm function to the product $\prod_{p\leq x} (1+a/p)$ we get $\sum_{p\leq x} \log(1+a/p) .$  Since $\log(1+x)\sim x$ when $x$ tends to $0$ then by Mertens theorem, we obtain $$\sum_{p\leq x} \log(1+a/p)\sim a \sum_{p\leq x} 1/p=a\log{\log{x}}+O(1)=\log{(\log{x})^a}-\log{(\log{2})^a}+O(1).$$ Then, we conclude that in the asymptotic limit ${x \rightarrow \infty}$, we have $$\prod_{p\leq x} (1+a/p)\leq\frac{(\log{x})^a}{(\log{2})^a}.$$ 
Is it true what I wrote?
Thanks in advance.
 A: I don't think it is correct. Note that if you have an asymptotic $$f\left(n\right)=g\left(n\right)+O\left(1\right)
 $$ it doesn't mean that $$f\left(n\right)\leq g\left(n\right)
 $$ for example $$\sum_{k\leq n}\frac{1}{k}=\log\left(n\right)+O\left(1\right)
 $$ but from the integral test we have $$\sum_{k\leq n}\frac{1}{k}\geq\int_{1}^{n+1}\frac{1}{t}dt=\log\left(n+1\right).
 $$ I think we can proceed as follow. Since $\log\left(1+x\right)<x
 $ we have $$\sum_{p\leq x}\log\left(1+\frac{a}{p}\right)\leq a\sum_{p\leq x}\frac{1}{p}.
 $$ Now we know that $$\sum_{p\leq x}\frac{1}{p}=\log\left(\log\left(x\right)\right)+M+O\left(\frac{1}{\log\left(x\right)}\right)\tag{1}
 $$ where $M$ is the Meissel–Mertens constant. From $\left(1\right)
 $ we have that exists some $A>0
 $ such that $$\left|\sum_{p\leq x}\frac{1}{p}-\log\left(\log\left(x\right)\right)-M\right|\leq\frac{A}{\log\left(x\right)}
 $$ and if $x$ is sufficiently large we have $$\sum_{p\leq x}\frac{1}{p}\leq\log\left(\log\left(x\right)\right)+M+\frac{A}{\log\left(x\right)}<\log\left(\log\left(x\right)\right)-\log\left(\log\left(2\right)\right)
 $$ since $M\approx0.2614
 $ and $-\log\left(\log\left(2\right)\right)\approx0.3665
 $. But this holds only for a sufficiently large $x$. If you want a result for all $x$ we can argue as follows. Again from $(1)$ exists some constant $C>0
 $ such that $$\sum_{p\leq x}\frac{1}{p}\leq\log\left(\log\left(x\right)\right)+C
 $$ for example in this paper (Wayback Machine) is shown that $$\left|\sum_{p\leq x}\frac{1}{p}-\log\left(\log\left(x\right)\right)\right|<6
 $$ if $x>e^{4}\approx54.598
 $. So if you want a result for all $x$ you have to find a constant that holds also for $x\in\left[2,54\right]
 .$ Hence we can conclude that $$\prod_{p\leq x}\left(1+\frac{a}{p}\right)\leq\log^{a}\left(x\right)e^{C},\, x\geq2$$ 
