Does the Inequality Involving Sums of Consecutive Twin Prime Pairs Always Hold?

[checkitagain]

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$$Let \ \ p_n \ \ = \ \ the \ \ nth \ \ prime \ \ number.$$Examples:$$p_1 \ = \ 2$$

$$p_2 \ = \ 3$$

$$p_3 \ = \ 5$$

$$p_4 \ = \ 7$$- - - - - - - - - - - - - - - - - - - - - - - - - - - - $$Let \ \ n \ \ belong \ \ to \ \ the \ \ set \ \ of \ \ positive \ \ integers.$$
Prove (or disprove) the following:$$p_n \ + \ p_{n + 1} \ \ \ge \ \ p_{n + 2} \ + \ p_{n - 2}, \ \ \ for \ \ all \ \ n \ \ge \ 4.$$Examples:$$\ \ 7 \ + \ 11 \ \ > \ \ 13 \ + \ \ 3$$

$$19 \ + \ 23 \ \ = \ \ 29 \ + \ 13$$

 

[CaptainBlack]

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$$Let \ \ p_n \ \ = \ \ the \ \ nth \ \ prime \ \ number.$$Examples:$$p_1 \ = \ 2$$

$$p_2 \ = \ 3$$

$$p_3 \ = \ 5$$

$$p_4 \ = \ 7$$- - - - - - - - - - - - - - - - - - - - - - - - - - - - $$Let \ \ n \ \ belong \ \ to \ \ the \ \ set \ \ of \ \ positive \ \ integers.$$
Prove (or disprove) the following:$$p_n \ + \ p_{n + 1} \ \ \ge \ \ p_{n + 2} \ + \ p_{n - 2}, \ \ \ for \ \ all \ \ n \ \ge \ 4.$$[/tex]Examples:$$\ \ 7 \ + \ 11 \ \ > \ \ 13 \ + \ \ 3$$

$$19 \ + \ 23 \ \ = \ \ 29 \ + \ 13$$

Asymtotically \(p_n+p_{n+1} \sim (2n+1)\log(x)\) and \(p_{n-2}+p_{n+2} \sim 2n\log(n)\).

so the inequality eventually holds, and how many terms we need to check explicitly before we can rely on the asymtotics can probably be determined from (on second thoughts it can't, the inequalities have too wide a spread):

\(n\log(n)+n\log(\log(n))-n<p_n<n\log(n)+n\log(\log(n)), \ \ n\ge 6\)

It appears to fail for \(n=29\), when \(p_n=109,\ p_{n+1}=113,\ p_{n-2}=103,\ p_{n+2}=127\)

There are also plenty of other counter examples for primes less than \(10^6\).

So, the new question is: Is there a \(n_0\) such that for all \(n>n_0\) the inequality holds? I would guess the answer is no.

CB