As the title ask, what is the best lower and upper bounds for the product below :
$$ \prod \limits\_{x < p \leq y} \frac{p+1}{p}$$
such that $p$ denote the prime numbers in which fullfil the conditions under the product ?



Best known bounds for a product over primes in an interval
Your product is the following $$\prod\_{y<p\leq x} \left(1+\frac{1}{p}\right).$$
You can use the fact that $$\prod\_{y<p\leq x} \left(1+\frac{1}{p}\right)\leq \prod\_{y<p\leq x} \left(1-\frac{1}{p}\right)^{-1} \leq \frac{\pi^2}{6} \prod\_{y<p\leq x} \left(1+\frac{1}{p}\right).$$ Note that $\frac{\pi^2}{6}=\frac{1}{\zeta(2)},$ where $\zeta$ is the Riemann zeta function. According to the best current explicit results of Pierre Dusart: For $x > 1$,
$$\prod\_{p\leq x} \left(1-\frac{1}{p}\right)< \frac{\exp(-\gamma)}{\ln{x}} \left(1+\frac{0.2}{\ln^3{x}}\right)$$
and for $x \geq 2278382,$ 
$$\prod\_{p\leq x} \left(1-\frac{1}{p}\right)> \frac{\exp(-\gamma)}{\ln{x}} \left(1-\frac{0.2}{\ln^3{x}}\right)$$
where $\gamma$ is the Euler's constant $(\gamma \sim 0.5772157)$ and you are done! 


