The lower bound for prime gaps

Question

Let $p_n$ denote the $n$-th consecutive prime number and $g_n=p_{n+1}-p_n$ a prime gap. There are many results about the upper bound for $g_n$. Some of them still has astatus of conjecture, such as Firoozbakth conjecture (in a prime gap version): $g_n<p_n\left( \sqrt[n]{p_n}-1\right) $ , $\forall n\in N$, and its consequence $g_n<\sqrt{n}$ , $n\geq 3645$ . Currently the best known proved result on upper bound is $g_n\leq p_n^{0.525}$ ( Baker-Harman-Pintz Theorem ). Heuristic and a few calculations that I made, suggest for upper bound:

Conjecture: $$\frac{g_n}{\log{g_n}}<2\log{n}, n\geq 5 $$

If we combine Conjecture with B.K.P. Theorem we get $g_n<1.05\log^2{n}$ , with Firoozbakth Conjecture we get $g_n<\log^2{n}$, which seem to be in contradiction with A.Granville proposition :$\limsup_{n\longrightarrow\infty}\frac{g_n}{\log^2{p_n}}\geq 2e^{-\gamma}\approx 1.1229$. There is also a simple consequence of Conjecture providing the lower bound for a prime gaps:
Proposition $$g_n>\left( \frac{p_{n+1}}{p_n}\right) ^{\frac{n}{2}} , n\geq 2$$
Proof: $$\left( \frac{p_{n+1}}{p_n}\right) ^{\frac{n}{2}}=\left[ \left( 1+\frac{g_n}{p_n}\right) ^{\frac{p_n}{g_n}}\right] ^{\frac{n.g_n}{2p_n}}<e^{\frac{n.g_n}{2p_n}}<e^{\log g_n}=g_n$$ ( follow from Conjecture and $\frac{p_n}{n}>\log n$) .

Question: Is conjecture plausible? Can it be proved or disproved? If Conjecture is not true, can we prove the Proposition , now taken as Conjecture, with other means?

Answer 1

1. The Proposition in the post is almost equivalent to the Conjecture, namely it implies that $$\frac{g_n}{\log g_n}\leq (2+o(1))\log n.$$ In particular, the Proposition (hence also the Conjecture) implies that $$g_n\ll\log n\,\log\log n.\tag{$\ast$}$$ This would contradict the common expectation that $g_n\gg (\log n)^2$ holds for infinitely many $n$'s.

2. On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to Ford-Green-Konyagin-Maynard-Tao (2014), and it states that $$g_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$ holds for infinitely many $n$'s.