[proofplan] The quantitative Chen lower-bound sieve estimate gives a positive lower bound for the integer $R_2(N)$. A positive value of this counting function means that at least one prime $p$ has $N-p$ equal to a $P_2$. [/proofplan] [step:Use the lower bound to get a positive count] Let $N$ be an even integer with $N\ge N_0$ and large enough that $\log N>0$. By the assumed Chen lower-bound sieve estimate, \begin{align*} R_2(N)\ge c\,\frac{N}{(\log N)^2}. \end{align*} The constant $c$ is positive, $N$ is positive, and $(\log N)^2$ is positive. Therefore \begin{align*} c\,\frac{N}{(\log N)^2}>0. \end{align*} It follows that \begin{align*} R_2(N)>0. \end{align*} [guided] The hypothesis gives the lower bound \begin{align*} R_2(N)\ge c\,\frac{N}{(\log N)^2}. \end{align*} For sufficiently large $N$, we have $\log N>0$. Since $c>0$ and $N>0$, the right-hand side satisfies \begin{align*} c\,\frac{N}{(\log N)^2}>0. \end{align*} Therefore the counting function is strictly positive: \begin{align*} R_2(N)>0. \end{align*} [/guided] [/step] [step:Extract the prime and almost-prime summand] The number $R_2(N)$ is a nonnegative integer counting primes $p<N$ for which $N-p$ is a $P_2$. Since $R_2(N)>0$, there exists at least one such prime $p$. Define \begin{align*} m:=N-p. \end{align*} By the defining property of the counted prime $p$, the integer $m$ is a $P_2$. Also \begin{align*} N=p+m. \end{align*} Thus $N$ has the required representation. [guided] Because $R_2(N)>0$ and $R_2(N)$ counts primes with a property, at least one prime $p<N$ is counted. The property of being counted is that $N-p$ is a $P_2$. Set \begin{align*} m:=N-p. \end{align*} Then $m$ is a $P_2$, and the definition of $m$ gives \begin{align*} N=p+m. \end{align*} This is exactly the desired representation with $p$ prime and $m$ a $P_2$. [/guided] [/step]