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

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... Jul 14 at 11:16
• Jul 14 at 11:20
• Evaluating $\prod_p (1±4/p^2)$ in Closed Form might be useful. Found using Approach0. Jul 14 at 11:31
• Where does this question come from? I've found this which suggests that this could be an open problem. Jul 14 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. Jul 14 at 11:44

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.