= Twin-prime upper bound from a quadratic sieve
{title2=$\#\{p\le x:p,p+2\text{ prime}\}\ll x/(\log x)^2$}
For $F(n)=n(n+2)$, every odd <prime> has two forbidden <residue classes>. Use <Selberg sieve weights> supported on odd <squarefree integers> $d\le R=x^{1/8}$. Their main quadratic form is $1/J$, where $J=\sum_{d\le R}\prod_{p\mid d}2/(p-2)$. The <truncated Euler-product lower bound> applied to primes up to $R^{1/8}$ gives $J\gg\log^2R$: the mean logarithmic divisor size is $2\sum_{3\le p\le R^{1/8}}(\log p)/p<\frac12\log R$, by the <Mertens first theorem>, and the full product is $\asymp\log^2R$, by the <Mertens second theorem>. The <optimal Selberg weights have modulus at most one>, so the remainder is at most $O(\sum_{d,e\le R}[d,e])=O(R^4)$. The finitely many possible pairs with $p\le R$ contribute $O(R)$. Thus the count is $O(x/J+R^4+R)$, proving the bound.
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