Show divergence of an infinite product with terms containing congruence-restricted primes

Question

Show that $\displaystyle\prod_{p\equiv1(3)}\frac p{p-1}=\infty$ where $p$ is a prime number.

I know that $\displaystyle\prod_p\frac p{p-1}=\infty$. However, how can I show the above one?

Is there any inequality that implies $\displaystyle\prod_p\frac p{p-1}<\displaystyle\prod_{p\equiv1(3)} \frac p{p-1}$?

Otherwise, can you help me to solve it?

Answer 1

We need two parts. The first part is $$\prod_{p\equiv1\bmod3}\frac p{p-1}=\prod_{p\equiv1\bmod6}\left(1+\frac1{p-1}\right)\ge\prod_{p\equiv1\bmod6}\left(1+\frac1p\right)\ge1+\sum_{p\equiv1\bmod6}\frac1p$$ The second part is proving the last infinite sum diverges. For one such proof, see here.

Therefore the original infinite product diverges.