# What is the upper bound for $\prod_{p|N}(1+p^{-1})$?

I'm learning Selburg Sieve in an Additive Number Theory book. Let $$N$$ be a positive even integer and $$r(N)$$ denote the number of primes $$p$$ less than $$N$$ such that $$N-p$$ is also a prime, in other words, $$r(n)$$ counts the representation of $$N$$ as a sum of two primes. Then $$r(N)=\mathcal O\bigg(\dfrac{N}{(\log N)^2}\prod_{p|N}\bigg(1+\dfrac 1p\bigg)\bigg).$$ The smaller terms as stated in the book are up to $$N^{9/10}$$, so I guess the order of $$\prod_{p|N}(1+p^{-1})$$ shouldn't exceed $$(\log N)^2$$, and then I have do some calculation on estimating the order:

Let $$N=p_1^{k_1}\cdots p_r^{k_r}$$ where each $$k_i\geq 1$$, then let $$M=p_1\cdots p_r$$, the "square-free kernel" of $$N$$, then the product is the same when calculating with respect to $$M$$:

\begin{align*}\prod_{p|N}\bigg(1+\dfrac1p\bigg)&=\prod_{p|M}\bigg(1+\dfrac1p\bigg)\\ &=\sum_{d|M}\dfrac1d\\ &\leq \sum_{d\leq M}\dfrac1d\\ &=\mathcal O(\log M)\\ &=\mathcal O(\log N).\end{align*} which fulfill my expectation. But is there any well-known smaller estimate about this product? For example I wish it could be estimated as $$\mathcal O(\log\log N)$$ or other function smaller than $$\log N$$.

(Sorry if this post is duplicated because I don't know the name of this product. )

Let $$f(N)=\prod_{p\mid N}(1+1/p)$$. For $$N=\prod_{1\leqslant i\leqslant r}p_i^{k_i}$$ (like above, $$p_i$$ are distinct primes and $$k_i>0$$), let $$m(N)$$ be the product of the first $$r$$ primes, and let $$l(N)$$ be the largest (i.e., $$r$$-th) of these primes.
From $$m(N)\leqslant N$$ and $$\lim\limits_{n\to\infty}(1/n)\sum_{p\leqslant n}\ln p=1$$ ($$\approx$$ PNT) we have $$l(N)=\mathcal{O}(\ln N)$$.
Now $$f(N)\leqslant f(m(N))=\mathcal{O}(\ln l(N))$$ because Mertens' results give $$\lim_{n\to\infty}\left[\ln n\prod_{p\leqslant n}\Big(1-\frac{1}{p}\Big)\right]=e^{-\gamma}\implies\lim_{n\to\infty}\left[\frac{1}{\ln n}\prod_{p\leqslant n}\Big(1+\frac{1}{p}\Big)\right]=\frac{6e^{\gamma}}{\pi^2}.$$ Finally, $$f(N)=\mathcal{O}(\ln\ln N)$$ as expected.