= Solution
The <Selberg upper-bound sieve> replaces the <indicator function> of a sifted set by the square of a divisor sum. If the real <Selberg sieve weights> satisfy $\lambda_1=1$, then
$$
\boldsymbol1_{(F(n),P(z))=1}
\le\left(\sum_{\substack{d\mid F(n)\\d\mid P(z)}}\lambda_d\right)^2,
\qquad P(z)=\prod_{p\le z}p.
$$
When $F(n)$ has no forbidden <prime factors>, the divisor sum is exactly one; otherwise its square is still nonnegative. Taking the weights to vanish for $d>z$ controls the error in summing this majorant. The main term is a positive <quadratic form> that can be diagonalized and minimized explicitly. This optimization, together with a bound for the remainder, is the central idea of the <Selberg sieve>.
For <twin primes>, take $F(n)=n(n+2)$ and let $\rho(d)$ be the number of its <roots> modulo a <squarefree integer> $d$. The <Chinese remainder theorem> makes $\rho$ a <multiplicative arithmetic function>, with $\rho(2)=1$ and $\rho(p)=2$ for odd <primes>. Thus the <polynomial root density in a sieve> is $g(d)=\rho(d)/d$, and
$$
\#\{n\le x:d\mid F(n)\}=xg(d)+r_d,\qquad |r_d|\le\rho(d).
$$
Summing the majorant gives
$$
S(x,z):=\#\{n\le x:(F(n),P(z))=1\}
\le xQ(\lambda)+O\left(\left(\sum_{d\le z}|\lambda_d|\rho(d)\right)^2\right),
$$
where $Q(\lambda)=\sum_{d,e\le z}\lambda_d\lambda_e g([d,e])$, with all indices <squarefree>. Here $\rho([d,e])\le\rho(d)\rho(e)$ controls the error.
We carry out the <Selberg sieve diagonalization>. Put
$$
h(r)=\prod_{p\mid r}\left(\frac1{g(p)}-1\right),\qquad
h(1)=1,\qquad
y_r=\sum_{\substack{d\le z\\r\mid d}}\lambda_dg(d).
$$
For <squarefree integers> $d,e$, <multiplicativity> gives
$$
g([d,e])=g(d)g(e)\sum_{r\mid(d,e)}h(r),
\qquad Q(\lambda)=\sum_{r\le z}h(r)y_r^2.
$$
The identity behind the first equality is $\prod_{p\mid(d,e)}[1+h(p)]=1/g((d,e))$. <Möbius inversion> gives $\lambda_1=\sum_r\mu(r)y_r=1$. The <Cauchy-Schwarz inequality> therefore implies
$$
1\le\left(\sum_{r\le z}h(r)y_r^2\right)
\left(\sum_{r\le z}\frac{\mu(r)^2}{h(r)}\right).
$$
Define $G(z)=\sum_{r\le z,\ r\text{ squarefree}}1/h(r)$. Equality is attained at $y_r=\mu(r)/(h(r)G(z))$, giving \b[$\min Q=1/G(z)$]. Inverting again, the optimizing <Selberg sieve weights> are
$$
\lambda_d=\frac{\mu(d)}{g(d)h(d)G(z)}G_d(z/d),\qquad
G_d(v)=\sum_{\substack{m\le v\\(m,d)=1\\m\text{ squarefree}}}\frac1{h(m)}.
$$
In particular $\lambda_1=1$. Since $G_d(z/d)\le G(z)$,
$$
|\lambda_d|\le\frac1{g(d)h(d)}
=\prod_{p\mid d}\frac1{1-g(p)}\le d.
$$
The last inequality holds separately at $p=2$, $p=3$, and $p\ge5$. Together with $\rho(d)\le d$, this crude bound already makes the remainder $O(z^6)$, because $\sum_{d\le z}d^2=O(z^3)$.
We also need a lower bound for $G(z)$; we prove it rather than presuppose a sieve asymptotic. For odd <squarefree integers> $r$,
$$
\frac1{h(r)}=\prod_{p\mid r}\frac2{p-2}\ge\frac{2^{\omega(r)}}r,
$$
where $\omega(r)$ counts distinct <prime factors>. Put $y=\sqrt z$ and $L(y)=\sum_{a\le y,\ a\text{ odd}}1/a=\tfrac12\log y+O(1)$. The total weight $1/(ab)$ of odd pairs $a,b\le y$ is $L(y)^2$. Exclude pairs for which $a$ has a square <prime factor>, $b$ has a square <prime factor>, or $a,b$ have a common <prime factor>. For each of these three categories the total weight is at most $L(y)^2\sum_{p\text{ odd}}p^{-2}$: for example, the weighted count of odd multiples of $p^2$ is $p^{-2}L(y/p^2)\le p^{-2}L(y)$, and the common-factor category uses $p^{-1}L(y/p)$ for each variable. Moreover,
$$
\sum_{p\text{ odd}}p^{-2}
\le\sum_{m\ge1}(2m+1)^{-2}
\le\frac19+\int_1^\infty\frac{dt}{(2t+1)^2}
=\frac5{18}.
$$
Hence the remaining odd <coprime> <squarefree> pairs have weight at least $L(y)^2/6$. Their products $r=ab\le z$ are <squarefree>, and $2^{\omega(r)}$ counts all ordered <coprime> factorizations of $r$. This proves the <elementary lower bound for the twin-prime sieve denominator>:
$$
G(z)\ge\sum_{\substack{r\le z\\r\text{ odd and squarefree}}}\frac{2^{\omega(r)}}r
\ge\frac{L(\sqrt z)^2}{6}\gg(\log z)^2.
$$
If $p>z$ and both $p,p+2$ are <primes>, $F(p)$ survives the sieve. The at most $z$ smaller values are the only exceptions. Thus, with $T(x)=\#\{p\le x:p,p+2\text{ prime}\}$ and $z=x^{1/10}$, the <twin-prime upper bound from a quadratic sieve> becomes
$$
T(x)\le S(x,z)+z\ll\frac{x}{(\log z)^2}+z^6+z,
\qquad\boxed{T(x)\ll\frac{x}{(\log x)^2}.}
$$
Finally <partial summation> gives
$$
\sum_{\substack{p\le X\\p,p+2\text{ prime}}}\frac1p
=\frac{T(X)}X+\int_2^X\frac{T(t)}{t^2}\,dt.
$$
The integral over $[3,\infty)$ is bounded by a constant times $\int_3^\infty dt/(t(\log t)^2)<\infty$. The nonnegative partial sums are therefore bounded and increasing. Thus \b[the reciprocal <series> over the smaller members of twin-prime pairs converges], a form of <Brun's theorem>. The upper bound does not assert that infinitely many <twin primes> exist.
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