Elementary lower bound for the twin-prime sieve denominator (${\sum }_{\begin{array}{c}d\le z\\ d\text{ squarefree}\end{array}}{\prod }_{p\mid d}\frac{\rho (p)}{p-\rho (p)}\gg (\log z{)}^2$)

For the polynomial $F(n)=n(n+2)$, take $\rho (2)=1$ and $\rho (p)=2$ at odd primes. On odd squarefree integers, the summand is at least $2^{\omega (d)}/d$. Put $y=\sqrt{z}$ and $L(y)={\sum }_{a\le y,a\text{ odd}}1/a=\frac{1}{2}\log y+O(1)$. The weight of odd pairs $a,b\le y$ is $L(y{)}^2$. The pairs for which $a$ is not squarefree, $b$ is not squarefree, or $(a,b)>1$ each have weight at most $L(y{)}^2{\sum }_{p\text{ odd}}{p}^{-2}$. Indeed, the weighted count of multiples of an odd $p^k$ is $p^{-k}L(y/p^k)\le p^{-k}L(y)$. Moreover,Hence the odd coprime squarefree pairs have weight at least $L(y{)}^2/6$. Their products $d=ab\le z$ are squarefree, and $2^{\omega (d)}$ counts all ordered coprime factorizations of $d$. This proves the displayed lower bound needed for the Selberg upper-bound sieve.