[proofplan] We compare prime pairs with sifted pairs. If both $n$ and $n+h$ are prime and $n>x^{1/2}$ and $n+h>x^{1/2}$, then neither integer has a prime divisor below $x^{1/2}$. The exceptional values with one of $n$ or $n+h$ at most $x^{1/2}$ contribute only $O_h(x^{1/2})$, which is absorbed by the stated upper bound for large $x$ and by changing the constant for bounded $x$. [/proofplan] [step:Embed prime pairs in the sifted set] Let $P_h(x)$ be the number of integers $n\le x$ such that $n$ and $n+h$ are both prime. Let $S_h(x)$ be the number of integers $n\le x$ such that neither $n$ nor $n+h$ has a prime factor below $x^{1/2}$. If $n>x^{1/2}$ and $n+h>x^{1/2}$ are both prime, then the only prime divisors of $n$ and $n+h$ are the numbers themselves. Hence neither has a prime divisor below $x^{1/2}$, so such an $n$ is counted by $S_h(x)$. The remaining prime-pair candidates satisfy $n\le x^{1/2}$ or $n+h\le x^{1/2}$. Since $h$ is fixed, there are $O_h(x^{1/2})$ such integers. Therefore \begin{align*} P_h(x)\le S_h(x)+O_h(x^{1/2}). \end{align*} [guided] Let $P_h(x)$ count integers $n\le x$ such that $n$ and $n+h$ are prime, and let $S_h(x)$ count integers $n\le x$ such that neither $n$ nor $n+h$ has a prime factor below $x^{1/2}$. If $n>x^{1/2}$ and $n+h>x^{1/2}$ are prime, then their only prime divisors are $n$ and $n+h$, so neither integer has a prime factor below $x^{1/2}$. Hence such an $n$ is counted by $S_h(x)$. The exceptions satisfy $n\le x^{1/2}$ or $n+h\le x^{1/2}$, and since $h$ is fixed there are $O_h(x^{1/2})$ exceptions. Therefore \begin{align*} P_h(x)\le S_h(x)+O_h(x^{1/2}). \end{align*} [/guided] [/step] [step:Apply the assumed upper-bound sieve] By the assumed dimension-two upper-bound sieve, \begin{align*} S_h(x)\ll_h \mathfrak S(\{0,h\})\frac{x}{(\log x)^2}. \end{align*} For $x$ sufficiently large, the elementary error $O_h(x^{1/2})$ is bounded by a constant multiple of $x(\log x)^{-2}$; after enlarging the implicit constant to handle bounded $x$, we obtain \begin{align*} P_h(x)\ll_h \mathfrak S(\{0,h\})\frac{x}{(\log x)^2}. \end{align*} This is the desired estimate. [guided] The analytic input is exactly the upper-bound sieve estimate for $S_h(x)$. The comparison step already proved that the prime-pair count is at most $S_h(x)$ plus $O_h(x^{1/2})$. Since $x^{1/2}=O(x(\log x)^{-2})$ for large $x$, this exceptional contribution is absorbed into the same upper-bound order. [/guided] [/step]