Let $p_n$ be the $n$th prime, for any integer $n$, prove that: $p_n+p_{n+1}\geq{p_{n+2}}$ I was just wondering about it. True or false, it seems a very interesting question to me. I'm also interested to see how this could be proven or disproven? Opinions are welcome as usual.
Regards
 A: Let $p_n$ denote the $n^{th}$ prime number. Bertrand's postulate states that for any $x \ge 1$, the interval $(x,2x)$ contains at least one prime number. Thus $p_{n+1} < 2 p_n$ for all $n \ge  1$.
Suppose to the contrary that there is an index $n$ for which $p_{n+2} > p_{n+1} + p_n$. This would imply that $p_{n+2} > \dfrac 32 p_{n+1}$.
A result related to Bertrand's postulate was proved by Nagura in 1952: if $x \ge 25$ then the interval $\left[x,\dfrac 65 x \right]$ contains at least one prime number. Thus if $p_{n+1} \ge 25$ we must have $$p_{n+2} \le \dfrac 65 (p_{n+1} + 1).$$
Combining these inequalities yields $\dfrac 32 p_{n+1} < \dfrac 65 (p_{n+1} + 1)$ which forces $p_{n+1} < 1$, a contradiction. Thus $p_{n+2} \le p_{n+1} + p_n$ as long as $p_{n+1} \ge 25$.
As for the remaining cases you have $5 \le 2+3$, $7 \le 3 + 5$, $11 \le 5 + 7$, and so on all the way to $29 < 23 + 19$.
A: Using the bounds $$n\log\left(n\right)+n\log\left(\log\left(n\right)\right)-1<p_{n}<n\log\left(n\right)+n\log\left(\log\left(n\right)\right),\, n\geq6$$
 your inequality holds if$$\left(n+2\right)\log\left(n+2\right)+\left(n+2\right)\log\left(\log\left(n+2\right)\right)<n\log\left(n\right)+n\log\left(\log\left(n\right)\right)+\left(n+1\right)\log\left(n+1\right)+\left(n+1\right)\log\left(\log\left(n+1\right)\right)-2$$
 and if rewrite$$n\left(\log\left(\frac{n+2}{n+1}\right)+\log\left(\frac{\log\left(n+2\right)}{\log\left(n+1\right)}\right)\right)+2\left(\log\left(n+2\right)+\log\left(\log\left(n+2\right)\right)\right)<n\log\left(n\right)+n\log\left(\log\left(n\right)\right)+\log\left(n+1\right)+\log\left(\log\left(n+1\right)\right)-2$$
 and now note that$$n\left(\log\left(\frac{n+2}{n+1}\right)+\log\left(\frac{\log\left(n+2\right)}{\log\left(n+1\right)}\right)\right)\longrightarrow1$$
 as $n\rightarrow\infty$
 so the LHS grow up like essentialy like $\log\left(n\right)$
  and the RHS grow up essentialy like $n\log\left(n\right).$
  So if $n$
  is sufficiently large is true.
