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 $p(i)\geq5$)?



Is this product, involving consecutive primes, always less than or equal to $1$?
We have [explicit bounds](https://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number)


$$ \log n + \log \log n - 1 < \dfrac{p(n)}{n} < \log n + \log \log n \ \text{for}\ n \ge 6 $$


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.


