Does this inequality always hold :
$$\frac{1}{6} \pi ^2 \prod \_{i=1}^x
 \frac{\left(p\_i\right){}^2-1}{\left(p\_i\right){}^2}\leq \frac{1}{p\_x}+1 $$


such that $p\_i$ is the $i$-th prime number



converge inequality for squares of primes
Yes. We have
$$
\frac{1}{6} \pi ^2 \prod \_{i=1}^x
 \frac{\left(p\_i\right){}^2-1}{\left(p\_i\right){}^2}=\prod\_{i>x} \frac{p\_i^2}{p\_i^2-1}\leqslant \prod\_{n=p\_x+1}^{\infty} \frac{n^2}{n^2-1}=\frac1{p\_x}+1.
$$


