On the inequality $\left(\frac{R_{n+1}}{R_n}\right)^nThe Wikipedia Firoozbakht's conjecture refers (see also the comments of OEIS A182514) an inequality due to Nicholson, I wondered if it is possible to prove the following conjecture.
Conjecture. The Ramanujan primes $R_n$ satisfy $$\left(\frac{R_{n+1}}{R_n}\right)^n<n^{5/4}(\log n)^3$$ for all integer $n>2$.
The corresponding Wikipedia for Ramanujan primes is this. My attempt of proof using the inequalities that I know was failed. I've tested the conjecture for the first few thousand of Ramanujan primes.

Question. Can you prove or refute previous conjecture? I'm asking for hints to prove it, or well feedback about its veracity: let's say what work can be done about it. Many thanks.

I've considered right add here one of those references from previous articles of Wikipedia.
References:
[1] Jonathan Sondow, Ramanujan primes and Bertrand's postulate, Amer. Math. Monthly, 116 (7), (2009), pp. 630–635.
 A: This answer ought to be understood in conjunction with the comments I have made on the above post: I am by no means an expert, but a conjecture like this, with no motivation or reason to exist, seems odd. Nevertheless, this conjecture is most likely true. With $P_n$ denoting the $n^{th}$ prime, we have  $P_{2n}<R_n<P_{4n}$ and, more interestingly, $R_n\sim P_{2n} $ as $n\to \infty$ (see OEIS or this paper). With this in mind, we can apply the prime number theorem to estimate
$$R_n\sim 2n\log(2n),$$
hence
$$\Big{(}\frac{R_{n+1}}{R_n}\Big{)}^n\sim\Big{(}\frac{2(n+1)\log(2(n+1)}{2n\log(2n)}\Big{)}^n.$$
As $n\to \infty$, this tends to $e$. It follows that the left hand side of the given conjecture therefore tends itself to a constant, and the result follows (trivially).
For fun, here is some Mathematica code, and a few graphs:
RamanujanPrime[n_] := 1 + Select[Range[4*n*Log[4 n]], PrimePi[#] - PrimePi[#/2] == n - 1 &]]//Max
RPApprox[n_, const_: 2] := const*n*(Log[const*n])

To see how good this approximation is, we run
DiscretePlot[{RPApprox[n], RamanujanPrime[n]}, {n, 1, 100}]

and obtain this:

To see OP's conjecture in action, we run
DiscretePlot[{(RamanujanPrime[n + 1]/RamanujanPrime[n])^n, n^(5/4)*Log[n]^3}, {n, 1, 100}]

(this could be made much slicker without recomputing the list, but oh well), and obtain this:

Often, it is not the answer itself that is interesting, but rather why the question was asked, and how the answer is obtained. It is worth keeping that in mind!
