Prove $\prod_p{(1-\frac{3}{p^2})}>\frac{1}{8}$

Question

Prove $$\prod_p{\left(1-\frac{3}{p^2}\right)}>\frac{1}{8},$$

where $p$ through out all prime numbers.

It' s equivalent to prove that $$ \sum_ {n = 1}^\infty \frac {3^{\Omega(n)}} {n^2} < 8,$$ where $\Omega(n)$ is the number of prime factor of n. For example, $\Omega(p^a)=a$.

The product is $\approx0.125487$ when $p＜100000$.

What is the background ?What have you tried ? It could be helpful for someone else... — Barackouda, Jul 14 '21 at 11:16
Evaluating $\prod_p (1±4/p^2)$ in Closed Form might be useful. Found using Approach0. — Toby Mak, Jul 14 '21 at 11:31
Where does this question come from? I've found this which suggests that this could be an open problem. — Toby Mak, Jul 14 '21 at 11:39
@ErikSatie Someone asked me that is there infinitely many positive integers n such that n-1,n,n+1 are all squarefree numbers, I found it's related to this product. — lsr314, Jul 14 '21 at 11:44

Aphelli · Accepted Answer · 2021-07-14T11:51:31.333

By taking the small primes away, it is enough to show that $\prod_{p \geq 7}{\left(1-\frac{3}{p^2}\right)} > \frac{75}{88}$.

If $p\geq 7$, $(1-3/p^2) \geq (1-1/p^2)^4$. Therefore, it is enough to show that $\prod_{p \geq 7}{\frac{1}{1-\frac{1}{p^2}}} < \left(\frac{88}{75}\right)^{1/4}$.

But expanding the product, we find that it is at most $1+\sum_{n \geq 49}{n^{-2}} \leq 1+1/48$. What remains to be shown is that $(1+1/48)^4 < 88/75$ and that is easy to check.

Prove $\prod_p{(1-\frac{3}{p^2})}>\frac{1}{8}$

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