Is this product, involving consecutive primes, always less than or equal to $1$? i don't know how to write math in Latex so i will try to explain it simply,
if we multiply 
$$\frac{p(i)^2}{p(i)^2-1}\prod_{j=1}^5\frac{p(i+j)^2-1}{p(i+j)^2}  ,$$ 
where $p(i)$ denote the $i$-th prime number, is this product always less or equal to 1 when $i\geq3$ (meaning $p(i)\geq5$)?
 A: We have explicit bounds
$$ \log n + \log \log n - 1 < \dfrac{p(n)}{n} < \log n + \log \log n \ \text{for}\ n \ge 6 $$
Let $b(n) = n \log n + n \log \log n$.
Thus for $n \ge 6$, your expression is less than
$$  B(i) = \dfrac{1}{1-1/(b(i)-i)^2}  \prod_{j=1}^5 \left(1 - \frac{1}{b(i+j)^2}\right) $$
It appears that $B(t) < 1$ for $t \ge 9$ (though rigourous bounds will be messy).  So (given that your expression is less than $1$ for $i=4,5,6,7,8$)
the answer is yes.
A: I checked with the  Maple software that for every $i\geq 3$, the mentioned formula is less than $1$, when $j=1 \cdots m$ where $m\geq2$. Please see the following picture:

A: Let $f(a)= 1+ 1/(a^2-1)$.  $f(a)$ decreases slowly to 1 as $a$ increases. If $p_j$ is a slowly increasing sequence of positive real numbers that increases slowly enough, then $\prod_{i=1}^5 f(p_{j+i}) \gt f(p_j)$.  How slowly? It suffices that $p_{j+5} \lt \sqrt{5}p_j$, which holds for the primes starting with $p_7=17$.  As a result, your inequality also holds for other sequences which grow faster than primes.
Gerhard "Generalization Is Good, Generally Speaking" Paseman, 2016.09.08
