Something weaker than Goldbach's Conjecture Inspired by this question on the "Sum of a odd prime and a odd semiprime as sum of two odd primes?", I wonder, if it is
possible to show, that every even number $2n\ge 12$, can be written as a sum of a prime and semiprime
$$p_1+p_2p_3=2n?$$
 A: Take a look at section 1, "Combinatorial Sieving", of this blog post of Terry Tao. He builds you up to the following weakening of the twin prime conjecture:

There are infinitely many $n$ such that both
$n$ and $n+2$ have the property that their every prime divisor is greater than $\sqrt[20]{n}$.

One can rewrite this into an even weaker statement:

There are infinitely $n$ such that both $n$ and $n+2$ are products of no more than $19$ primes

Roughly, the proof is as follows: Let $z = \sqrt[20]{N}$. Heuristically, for $n$ odd, the probability that neither $n$ nor $n+2$ is divisible by a prime less than $z$ is $\prod_{3 \leq p < z} (1-2/p) \approx c/(\log z)^2$ where the value of the constant $c$ is unimportant. So one would expect that there are about $c N/(\log z)^2$ integers less than $N$ meeting the given conditions. If you try to make this argument rigorous directly, you get huge error terms, but if you are very careful you can make it work. See Terry's post for the careful details.
I don't have a big picture understanding of why you couldn't use the same argument to actually prove the twin prime theorem, taking the product $\prod_{3 \leq p < \sqrt{N}} (1-2/p)$ instead of $\prod_{3 \leq p < z} (1-2/p)$. It's just that when you get into the nitty gritty of the proof, $\sqrt[20]{N}$ is small enough and $\sqrt{N}$ isn't.
I have never looked at the proof of Chen's theorem, but sieving methods refer to arguments like this one, so you might want to read this example of a sieving proof (already quite challenging, in my opinion), before taking on what Eisenstein describes as a "highly technical application of sieving methods".
