The idea of this post arises when I've considered simple variants of the known as *Firoozbakht's conjecture* (see this corresponding [Wikipedia](https://en.wikipedia.org/wiki/Firoozbakht%27s_conjecture) or [1]), and comparisons by trial and error with means (if you are interested feel free to do yourself comparisons motivated from the Wikipedia *Generalized mean*).


I've considered the variant, that is a comparison,
 $$\frac{n+p\_{n+1}}{n+1}>\lambda(n)\cdot p\_{n}^{\frac{1}{n}}+\mu(n)\tag{1}$$
where $p\_k$ denotes the $k$-th prime number and $\lambda(n)$ and $\mu(n)$ are simple functions. As I've said in the introductory paragraph I got my comparison by trial and error using a Pari/GP program, from these calculations I've considered $\lambda(n)=\log n$ and $\mu(n)=\log\log n$, and my proposal (below in my **Question**) will be for integers $n\geq 2700$.


The identity $(1)$ evokes a comparison for means because the quantity $\frac{n\cdot 1+p\_{n+1}}{n+1}$ is the arithmetic mean of $n$ ones $1$'s and $p\_{n+1}$ while that $p\_{n}^{1/n}$ is the geometric mean of $n-1$ ones $1$'s and the prime $p\_n$. The functions $\log n$ and $\log\log n$ are simple functions that additionally belong to a class of functions studied in Rafael Jakimczuk, *Functions of Slow Increase and Integer Sequences*, Journal of Integer Sequences, Article 10.1.1, Vol. 13 (2010).



> 
> **Question.** Prove or refute the following conjecture as a variant of Firoozbakht's conjecture: 
> 
> 
> *For each integer* $n\geq 2700$ *the following inequality*
> $$\frac{n+p\_{n+1}}{n+1}>p\_{n}^{\frac{1}{n}}(\log n)+\log\log n\tag{2}$$
> *holds.* 
> 
> 
> **Many thanks.**
> 
> 
> 


My proposal $(2)$ is true for integers $2700\leq n\leq 50000$. You can see it from the web Sage Cell Server using this Pari/GP program (choose *GP* as language) 


`for(n=2700, 50000, if((n+prime(n+1))/(n+1)<(log(n))*prime(n)^(1/n)+log(log(n)),print(n)))`


(you need to wait a minute to see that there aren't counterexamples as outputs of this program), or this other output, it is a line written in Pari/GP, showing the differences of LHS and RHS 


`for(n=2700, 50000, if((n+prime(n+1))/(n+1)>(log(n))*prime(n)^(1/n)+log(log(n)),print((n+prime(n+1))/(n+1)-(log(n))*prime(n)^(1/n)-log(log(n)))))`


As was said feel free to do yourself variants in your home if you can to create a more interesting variant of Firoozbakht's conjecture using generalized means. 


References:
-----------


[1] *Conjecture 30. The Firoozbakht Conjecture*, Retrieved 22 August 2012 in Carlos Rivera's web The Prime Puzzles & Problems Connection.



On an attempt to create interesting variants of Firoozbakht's conjecture, evoking combinations of different generalized means
Yes, (2) holds for all large enough $n$.
According to Wikipedia (see also [this question](https://mathoverflow.net/q/259997/18698)) $\frac{p\_n}n=\log n+\log\log n-1+\frac{\log\log n}{\log n}(1+o(1))$, so
$$\frac{n+p\_{n+1}}{n+1}=\log(n+1)+\log\log(n+1)+\frac{\log\log n}{\log n}(1+o(1))$$
Since $\log\log(n+1)\approx\log\log n+\frac1{n\log n}$, so that $\log\log(n+1)-\log\log n=o\left(\frac{\log\log n}{\log n}\right)$, (2) now follows from:



> 
> Lemma : $\log(n+1)-p\_n^{1/n}\log n=o\left(\frac{\log\log n}{\log
> n}\right)$.
> 
> 
> 


Proof: Using the prime number theorem to write $p\_n=n\log n\cdot(1+o(1))$ we have $p\_n^{1/n}=\exp\left(\frac1n(\log n+\log\log n+o(1))\right)=1+o\left(\frac{\log\log n}{(\log n)^2}\right)$. On the other hand, $\log(n+1)=\log n+o\left(\frac{\log\log n}{\log n}\right)$ and the lemma follows.


