I'm trying to prove - without any success - the following result.
Let $y$ be a positive integer and let $r:=\pi(y)-1$, the number of primes not exceeding $y$ (but not counting $2$). Let $p_1,\dots,p_r$ be all the odd primes not exceeding $y$. For $f=1,...,r$ let's define $$s_f:=\sum_{1\leq i_1<\dots<i_f\leq r}{\frac{2^f}{p_{i_1}\cdots p_{i_f}}}$$ I would like to prove that $s_f \geq s_{f+1}$ for all $f$. I do not have any problem in showing this result for $f \geq r/2$, because in this case the number of summands is decreasing. I only have problem to prove it for all $f<r/2$.
I really hope in your help.