**Inequality**


What values of $n$ satisfy the following inequality?



> 
> $$2(n-2) < Ap\_n\prod\_{i=3}^n \left(\frac{p\_i-2}{p\_i}\right)$$
> 
> 
> 


$p$ are prime numbers and the notation $p\_i$ indicates the $i$th prime number.


A comes from the relationship $p\_n = 6A + r : 0 \leq r < 6$


(This is a slight variation of [The values of $n$ which satisfy an inequality about prime numbers](https://mathoverflow.net/questions/234793/the-values-of-n-which-satisfy-an-inequality-about-prime-numbers))



Find the values of $n$ that satisfy this inequality involving a product over prime numbers
We already have $p\_n \gt 2(n-2)$ for $n \gt 0$ by elementary (non-analytic) methods. Similar methods can establish the inequality for sufficiently large n. 


Suppose the inequality $1 \lt A \prod\_{i=3}^n \frac{p\_i - 2}{p\_i}$ holds for positive integers $n=j$ and $j+1$. I will show the inequality holds for $j+2$, leaving the inductive conclusion to the reader. Note that the term for $j+2$ is at least the term for $j$ times $B=\frac{(A+1)(p\_{j+1}-2)(p\_{j+2}-2)}{Ap\_{j+1}p\_{j+2}}$. This is because $p\_{j+2} \geq 6 + p\_j$, so the value of $A$ for the two terms increases by at least 1. But $\frac{A+1}{A}\geq 1 + 6/p\_j$ while $(1 - 2/p\_{j+1})(1- 2/p\_{j+2})\gt 1 - 4/p\_j)$ giving $B \gt 1$ and thus the inequality holds for $j+2$.


One can generalize this replacing $A$ by smaller values like $cp\_n$ for some positive real $c$ and replacing $p\_i-2$ by $p\_i-k$ for a larger fixed value of $k$. The rewritten inequality will fail for small $n$ but eventually hold, as analytic methods will have the right hand side grow like a constant times $p\_n^2/\log p\_n$.


Gerhard "Should Anticipate The Next Questions" Paseman, 2016.03.30.


