How to prove this asymptotic formula?

Question

How to prove this asymptotic formula?

$$ \prod\limits_{p\leq x}\left(1+\frac{1}{p}\right) \sim \frac{6 e^C}{\pi^2}\log x $$

Where we multiply over all primes less than or equal to x.

I have little idea on how to approach this. My first thought was to write $ \prod\limits_{p\leq x} \left(1-\frac{1}{p^2}\right) = \prod\limits_{p\leq x}\left(1-\frac{1}{p}\right) \prod\limits_{p\leq x} \left(1+\frac{1}{p}\right)$, but I'm stuck there.

Answer 1

Just to put things together, we know that: $$\prod_{p}\left(1-\frac{1}{p^2}\right) = \frac{1}{\zeta(2)}=\frac{6}{\pi^2}\tag{1}$$ while Mertens' third theorem gives: $$\prod_{p\leq n}\left(1-\frac{1}{p}\right)\sim\frac{e^{-\gamma}}{\log n}.\tag{2}$$

Answer 2

Here's a solution. By Mertens we have: $\prod\limits_{p\leq x} \left( 1 - \frac{1}{p} \right) \sim \frac{1}{e^\gamma \log x}$

Then combining as suggested by Daniel, we obtain:

$$ \prod\limits_{p≤x}\left(1 + \frac{1}{p} \right) = \frac{\prod\limits_{p≤x}\left(1 - \frac{1}{p^2} \right)}{\prod\limits_{p≤x}\left(1 - \frac{1}{p} \right)} \sim \frac{\frac{1}{\zeta(2)}}{\frac{1}{e^\gamma \log x}} = \frac{6 e^\gamma}{\pi^2} \log x $$