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 showing the differences of LHS and RHS 

`for(n=2700, 50000, if((n+prime(n+1))/(n+1)>prime(n)^(1/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:
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[1] *Conjecture 30. The Firoozbakht Conjecture*, Retrieved 22 August 2012 in Carlos Rivera's web The Prime Puzzles & Problems Connection.