Primes in a concave sequence

Question

Are there infinitely many consecutive tuples of primes $(p_{n-1},p_n,p_{n+1})$ satisfying $$\frac{1}{2}(p_{n-1}+p_{n+1})\le p_n$$ I don't know how to do this problem. Maybe we use the prime number theorem? Thanks for all the help. My friend has found by checking that probably this guess is true. But no rigorous proof of it yet.

Answer 1

Yes there are. For suppose not. Then $$p_{n+1}-p_n\le p_n-p_{n-1}$$ has only finitely many solutions. That is, from some point on we have always $$p_{n+1}-p_n>p_n-p_{n-1}\ .$$ So $$p_{n+1}>p_n+1\ ,\quad p_{n+2}>p_n+1+2\ ,\quad p_{n+3}>p_n+1+2+3$$ and so on. This means, essentially, that $p_k$ is at least a constant times $k^2$. But this is not true, for it is a consequence of the Prime Number Theorem that $p_k$ is asymptotically $k\log k$.