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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 $p<100000$.

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    $\begingroup$ What is the background ?What have you tried ? It could be helpful for someone else... $\endgroup$
    – Erik Satie
    Jul 14 at 11:16
  • $\begingroup$ See en.wikipedia.org/wiki/Bonse%27s_inequality $\endgroup$
    – Erik Satie
    Jul 14 at 11:20
  • $\begingroup$ Evaluating $\prod_p (1±4/p^2)$ in Closed Form might be useful. Found using Approach0. $\endgroup$
    – Toby Mak
    Jul 14 at 11:31
  • $\begingroup$ Where does this question come from? I've found this which suggests that this could be an open problem. $\endgroup$
    – Toby Mak
    Jul 14 at 11:39
  • $\begingroup$ @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. $\endgroup$
    – lsr314
    Jul 14 at 11:44
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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.

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