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
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
Yes. We have $$ \frac{1}{6} \pi ^2 \prod _{i=1}^x \frac{\left(p_i\right){}^2-1}{\left(p_i\right){}^2}=\prod_{i>x} \frac{p_i^2}{p_i^2-1}\leqslant \prod_{n=p_x+1}^{\infty} \frac{n^2}{n^2-1}=\frac1{p_x}+1. $$