Firoozbakht's conjecture states that for all $n\geq 1$
$$p_n^{\frac{1}{n}}>p_{n+1}^{\frac{1}{n+1}},$$
where $p_k$ the kth prime number. By asumption of this conjecture, for a fixed $n$, there is a positive integer $a_n$ such that
$$p_n^{n+1}=a_n+p_{n+1}^n,$$
and combining with abc conjecture, since $gcd(a_n,p_{n+1}^n,p_n^{n+1})=1$, $\forall \epsilon>0$ $\exists \mu(\epsilon)>0$ such that
$$p_n^{n-2}<\mu(\epsilon)rad(a_n)^{1+\epsilon}p_n^{3\epsilon},$$
where $rad(m)=\prod_{\substack{p\mid n,\text{p, prime}}}p$, if $m>1$ and $rad(1)=1$. I've used Firoozbakht's conjecture for such deduction and poor computations. From $p_n^{n+1}=a_n+p_{n+1}^n$, one has unconditionally that
$$p_n^{n+1}-p_{n}^n>a_n,$$
thus combining with the obvious fact that $rad(m)\leq m$, one has
$$rad(a_n)<p_n^{n+1}-p_{n}^n.$$ But i know that this computation is poor too.
Question. Can you improve my computations looking for a reasonable comparision between the Firoozbakht's conjecture and the abc conjecture? I say that you know improve my computations and know how combine with a more best strategy this statement for previous triples $(a_n,p_{n+1}^n,p_n^{n+1})$. What about $rad(a_n)$? I say if you can improve my bound using the equation for previous triples. Thanks in advance.