Can we prove: let $2<p<q$ be two consequtive prime numbers, then there is always a prime number in interval $(q,p+q]$ (ie. betwen $q$ and $p+q$). In other words: if $p\_{n-1}<p\_n< p\_{n+1}$ are any three consequtive primes, then $p\_{n+1}<p\_{n}+p\_{n-1}$. Bertrand postulate is a consequence of this statement, since $p+q<2q$



Bertrand postulate- stronger version
By Section 4 of P. Dusart [[Math. Comp. 68(1999), 411--415](https://www.ams.org/journals/mcom/1999-68-225/S0025-5718-99-01037-6/S0025-5718-99-01037-6.pdf)], for any $x\ge3275$ there is a prime $p$ such that
$$x\le p\le x\left(1+\frac1{2\log^2x}\right)\le x\left(1+\frac1{2\log^2 3275}\right)<1.01x.$$


For any integer $n\ge463$, we have $p\_n>3275$ and hence
$$p\_{n+1}<1.01p\_n<p\_n+\frac12p\_n<p\_n+p\_{n-1}.$$


