Let us observe the following pattern
$N - p_1 = m_1, N - p_2 = m_2, \ldots , N - p_r = m_r$; take $p_1 = 3$ and $p_2 = 5,\ldots$ notice that $p_r$ is the larger prime less than or equal to square root of $N$. Here $N$ is an even number and large, where as $m_1,\ldots,m_r$ are composite numbers. What I guessed, with reference to above data all even numbers smaller than $N$ can be written by two primes and then we assume that all $m_1,\ldots,m_r$ are composite and we will face an absurdity (contradiction). Can we generalize my guess, if I am not wrong.
As per the request, I am adding more information on this post.
In my previous post I have discussed the same. But, Once again I am giving below.
Consider $m_1$...$m_r$ less than or equal to $m_1$.$a_1$.$a_2$...up to a_(r-1), where $a_1$ to a_r-1 are composite numbers smaller than $m_1$ and may be equal or larger than $m_2$ to $m_r$ by this information we conclude the following.
1) Take prime factors of K and restrict the value of K, where K belongs to [3, square root of $m_1$] will appeared in {$m_1$,$a_1$,$a_2$,...up to a_ r-1}.
2) If L to the member of K belongs to [3, square root of N] are in equation N - $p_i$ less than or equal to tK, where i belongs to [1, r] and L = r/K or < r/K, for all K belongs to [3, square root of N]
3) If M to be the number of S > square root of N are in N - $p_i$ less than or equal to $S_q$ that these numbers are odd and i belongs to [1, r].
4) also, we always have; log (N-3) ...(N-$p_r$) > r log(N-$p_r$)
I hope by the above in formation we can conclude the goldbach conjecture, if I am not wrong..
