I think it could be a safe assumption that this is not equivalent to RH in a simple way (assuming the other assertions of the question are true).
Here is why: to show that RH implies Goldbach (at least asymptotically) is not at all an unnatural idea, which however as far as I know is open.
For example, in 'Refinements of Goldbach's Conjecture, and the
Generalized Riemann Hypothesis' Granville discusses questions close to this.
However, it seems to me that the asymptotic counts of the number of solutions to 'the Goldbach equations' are related to the RH (and GRH).
Another example would be Deshouillers, Effinger, te Riele, Zinoviev 'A complete Vinogradov $3$-primes theorem under the Riemann hypothesis. Electron. Res. Announc. Amer. Math. Soc. 3 (1997), 99–104.' who showed that ternary Goldbach follows from GRH.
This is less directly related as ternary Goldbach is long known asymptotically, and this is thus about eliminating 'small' counterexamples.
So, it just seems more than a bit unlikely that 'RH implies asymptotic Goldbach' can be solved with a half-page argument and an equivalence argument direct enough that somebody might simply supply it here.
In addition, despite an explicit request (made a while ago) there is still no information/evidence provided why this should be equivalent to RH, which I interpret as the absence of any such evidence.
Finally, since being equivalent is a bit of a vague notion (if both were true they were equivalent even if totally unrelated) and since this is too long for a comment anyway, I thought I give these generalities as an answer.
The major difference is the choice of the Selberg Sieve weights. I strongly recommend reading Maynard's paper. It is well written, and the core ideas are nicely explained.
In what follows, I'll give a brief explanation of what Selberg Sieve weights are, why they appear, and what was chosen differently in Maynard's paper compared to that of Goldston Pintz and Yildirim.
Let $\chi_{\mathcal{P}}(n)$ denote the indicator function for the primes. Given an admissible $k$-tuple $\mathcal{H}=\{h_1<h_2<\cdots<h_k\}$ the goal is to choose a non-negative function $a(n)$ so that
$$\sum_{x<n\leq 2x} \left(\sum_{i=1}^k \chi_{\mathcal{P}}(n)(n+h_i)\right)a(n)\geq \rho \sum_{x<n\leq 2x} a(n)\ \ \ \ \ \ \ \ \ \ (1)$$
for some constant $\rho>1$. If this is the case, then by the non-negativity of $a(n)$, we have that for some $n$ between $x$ and $2x$ $$\left(\sum_{i=1}^k\chi_{\mathcal{P}}(n)(n+h_i)\right)\geq \lceil\rho \rceil,$$ and so there are at least $\lceil\rho \rceil$ primes in an interval of length at most $h_k-h_1$. This would imply Zhangs theorem on bounded gaps between primes, and if we can take $\rho$ arbitrarily large, it would yield the Maynard-Tao theorem. However, choosing a function $a(n)$ for which equation $(1)$ this holds is very difficult. Of course we would like to take $$a(n)=\begin{cases}
1 & \text{when each of }n+h_{i}\text{ are prime}\\
0 & \text{otherwise}
\end{cases} ,$$
as this maximizes the ratio of the two sides, but then we cannot evaluate $\sum_{x<n\leq 2x} a(n)$ as that is equivalent to understanding the original problem. Goldston Pintz and Yildirim use what is known as Selberg sieve weights. They chose $a(n)$ so that it is $1$ when each of $n+h_i$ are prime, and non-negative elsewhere, in the following way:
Let $\lambda_1=1$ and $\lambda_d=0$ when $d$ is large, say $d>R$ where $R<x$ will be chosen to depend on $x$. Then set $$a(n)=\left(\sum_{d|(n+h_1)\cdots(n+h_k)} \lambda_d\right)^2.$$ By choosing $a(n)$ to be the square of the sum, it is automatically non-negative. Next, note that if each of $n+h_i$ is prime, then since they are all greater than $x$, which is greater than $R$, the sum over $\lambda_d$ will consist only of $\lambda_1$. This means that $a(n)=1$ when each of the $n+h_i$ are prime, and hence $a(n)$ satisfies both of the desired properties. Of course, we do not want $a(n)$ to be large when none, or even very few of the $n+h_i$ are prime, so we are left with the difficult optimization problem of choosing $\lambda_d$. The optimization in the classical application of the Selberg sieve to bound from above the number of prime-tuples involves diagonalizing a quadratic form, leading to
$$\lambda_d = \mu(d) \left(\frac{\log(R/d)}{\log R}\right)^k,$$
and so
$$a(n)=\left(\sum_{\begin{array}{c}
d|(n+h_{1})\cdots(n+h_{k})\\
d<R
\end{array}}\mu(d)\left(\frac{\log(R/d)}{\log R}\right)^{k}\right)^{2}.$$
This choice is not so surprising since the related function $\sum_{d|n}\mu(d) \left(\log(n/d)\right)^k$ is supported on the set of integers with $k$ prime factors or less. Goldston Pintz and Yildirim significantly modified this, and used a higher dimensional sieve to obtain their results.
The critical difference in the approach of Maynard and Tao is choosing weights of the form
$$a(n)=\left(\sum_{d_{i}|(n+h_{i})\ \forall i}\lambda_{d_{1}\cdots d_{k}}\right)^{2}.$$ This gives extra flexibility since each $\lambda_{d_1,\dots,d_k}$ is allowed depend on the divisors individually. Using this additional flexibility Maynard and Tao independently established equation $(1)$ for any $\rho>1$, with $k$ depending on $\rho$.
Here are four great resources where you can read more:
- Maynard's paper.
- Tao's blog posts.
- Granville's article on the Maynard-Tao theorem and the work of Zhang.
- Soundararajan's exposition of Goldston Pintz and Yildirim's argument.
Best Answer
As far as I know there are two approaches to Goldbach type problems, the circle method and sieve methods. In the sequel I will restrict myself to the circle method, hoping that someone else writes something about sieves.
Define the exponential sum $S(\alpha)=\sum_{n\leq N}\Lambda(n) e(n\alpha)$, where $e(x)=e^{2\pi i x}$. By orthogonality of $e(x)$ we have that the (weighted) number of representations of $N$ as the sum of 3 primes is $$ \int_0^1 S(\alpha)^3e(-N\alpha)\;d\alpha. $$ If $\alpha=\frac{p}{q}$ is rational, then $$ S(\alpha)=\sum_{a=0} ^{q-1} e \left( \frac{ap}{q} \right) \underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n), $$ which can be approximated using the prime number theorem for arithmetic progressions. If $\alpha$ is very close to a rational number, evaluation can still be done by partial summation. Hence define the major arcs $\mathfrak{M}\subseteq[0,1]$ as the set of integers very close to rational numbers with small denominator. Then $$ \int_\mathfrak{M}S(\alpha)^3e(-N\alpha)\;d\alpha $$ can approximately be evaluated, it turns out that this integral essentially equals the expected main term. Let $\mathfrak{m}=[0,1]\setminus\mathfrak{M}$ be the so called minor arcs. To prove that every sufficiently large odd integer is the sum of three primes, Vinogradoff showed that for suitably defined arcs the integral over the major arcs is $\sim\mathfrak{S}(N)N^2$, and that for $\alpha\in\mathfrak{m}$ we have $|S(\alpha)|\ll\frac{N}{\log^2 N}$. Hence \begin{eqnarray*} \int_\mathfrak{m}S(\alpha)^3e(-N\alpha)\;d\alpha & \leq &\int_\mathfrak{m}|S(\alpha)^2|\;d\alpha\max_{\alpha\in\mathfrak{m}}|S(\alpha)|\\ & \leq & \int_0^1|S(\alpha)^2|\;d\alpha\max_{\alpha\in\mathfrak{m}}|S(\alpha)|\\ & = & \sum_{n\leq N}\Lambda(n)^2\max_{\alpha\in\mathfrak{m}}|S(\alpha)|\\ & \ll & \frac{N^2}{\log N} \end{eqnarray*} So for large $N$ the error coming form $\mathfrak{m}$ is of smaller order then the main term, and we obtain an asymptotic for the number of representations. Since proper powers are pretty rare, passing from $\Lambda$ to primes is no problem.
All this was known and used by Hardy and Littlewood some 20 years before Vinogradov, however, they could not give an unconditional bound for $S(\alpha)$, and had to assume the generalized Riemann hypothesis. Among the things Helfgott did, was a numerical bound which is non-trivial for rather small values of $N$, which in analytic number theory is usually pretty difficult. However, no matter what progress may come, the argument used for odd $N$ can never prove binary Goldbach. The crucial point in the argument above was that the integral over the major arcs is asymptotically equal to the conjectural main term, and the integral over the minor arcs is of smaller order. The integral over the major arcs can still be evaluated, and is $\mathfrak{S}(N)N$, as it should, however, $$ \int_0^1 |S(\alpha)|^2\;d\alpha = \sum_{n\leq N}\Lambda(n)^2\sim N\log N $$ is bigger then the main term. To prove Goldbach for even integers we have to show that $\left|\int_\mathfrak{m} S(\alpha)^2e(-N\alpha)\;d\alpha\right|$ is considerably smaller than $\int_\mathfrak{m} |S(\alpha^2)|\;d\alpha$, that is, that there is some cancelation within the integral.
Still, one can obtain interesting results in this way. For example, Montgomery and Vaughan (The exceptional set in Goldbach's problem, Acta Arith. 27) have shown that all even integers up to $x$ with at most $x^{1-c}$ exceptions can be written as the sum of two primes. They used the fact that $\int_\mathfrak{m} S(\alpha)^2e(-N\alpha)\;d\alpha$ cannot simultaneously be large for many different $N$.
Somewhat surprisingly Brüdern, Granville, Perelli, Vaughan, and Wooley, (On the exponential sum over k-free numbers, R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 356) showed that there are interesting binary additive problems which can be solved by the circle method, e.g. problems involving squarefree integers. However, as they explicitly mention, their method fails for sets of density 0, the reason being the same as above.