I was inspired by this post to look at some prime numbers and see if I can find a bound.
In the post, a conjecture was made with variables $m$ and $n$ such that $(m,n)\neq (1,1)$ respectively. I was able to prove that if it was true, then $m=n$ would imply a tighter upper bound for $p_{n+1}$. A current upper bound for $p_{n+1}$ is $2p_n$, but the OP's conjecture would imply that $2p_{n+1}<p_{2n+1}$ which is tighter and better than the inequality, $p_{n+1}<2p_n$.
Doing some tests, I thought that a better inequality would be that $2p_{n+1}\le p_{2n+1}-1$, but then I decided to do some more tests and I came up with a conjecture.
Conjecture:
$$\begin{align}\frac{p_{2n+1}-1}{2}&\geqslant p_{n+1}+2(n+1)^{2/(n-1)}\underbrace{+\,\frac 12\log_4^2(n-6)}_{\text{recent edit}}\tag{$n>6$} \\ \text{with equal}&\text{ity iff $n=7$.}\end{align}$$
Can this be proven/disproven?
Thank you in advance.
Edit: For a tighter bound, I added the term ${}^1/{}_2\log_{4}^{2}(n-6)^\dagger$ to the RHS (Right-Hand Side).
${}^\dagger$By $\log_b^2(x)$, I mean $\big(\log_b(x)\big)^2$.
Using equations $(11)$ and $(12)$ from here, for $n\ge688683$, $$\frac12p_{2n+1}-p_{n+1}>f(n)$$ where $$f(n)=\frac{2n+1}2\left(\ln(2n+1)+\ln\ln(2n+1)-1-\frac{\ln\ln(2n+1)-2.1}{\ln(2n+1)}\right)-(n+1)\left(\ln(n+1)+\ln\ln(n+1)-1-\frac{\ln\ln(n+1)-2}{\ln(n+1)}\right)$$ and clearly $f(n)-\left(\frac{1}{2}+2(n+1)^{\frac{2}{n-1}}+\frac{\ln^2(n-6)}{\ln^24}\right)$ is monotonically increasing as the $x\ln x$ terms in $f$ dominate. It can be computationally verified that the inequality holds for $n<688683$.