Prove: If $P_n$ is $n$-$th$ prime number then $P_{n+m} \ge P_n+P_m$

Question

Let $n > 1$ and $m > 0$ be two integers and $P_n$ be the $n^{th}$ prime.

Prove: $$P_{n+m} \ge P_n + P_m .$$

Can you give a hint, reference, comment, or proof?

Answer 1

This is an expanded version of my previous answer. It shows, among other things, that the OP's conjecture contradicts the $k$-tuple conjecture for $k=459$.

1. Let $r\geq 0$ be a fixed integer. I claim that the following two statements are equivalent (for integral variables $x,y,n,m$): $$\forall x,y\geq 2:\pi(x+y)\leq\pi(x)+\pi(y)+r\tag{1}$$ $$\forall n,m\geq 2: P_n+P_m-1\leq P_{n+m+r-1}\tag{2}$$

$(1)\Rightarrow(2)$: Let $n,m\geq 2$ be arbitrary, and apply $(1)$ for $x:=P_n-1$ and $y:=P_m-1$, which are at least $2$. We get $\pi(P_n+P_m-2)\leq\pi(P_n-1)+\pi(P_m-1)+r=n+m+r-2$, whence $P_n+P_m-2<P_{n+m+r-1}$. That is, $P_n+P_m-1\leq P_{n+m+r-1}$, which is $(2)$.

$(2)\Rightarrow(1)$: Let $x,y\geq 2$ be arbitrary, and apply $(2)$ for $n:=\pi(x)+1$ and $m:=\pi(y)+1$, which are at least $2$. We get $x+y\leq P_n+P_m-2\leq P_{n+m+r-1}-1$, whence $\pi(x+y)\leq n+m+r-2$. That is, $\pi(x+y)\leq\pi(x)+\pi(y)+r$, which is $(1)$.

2. The OP's conjecture is weaker than $(1)$ and $(2)$ for $r=0$, but stronger than $(1)$ and $(2)$ for $r=1$. At any rate, the $k$-tuple conjecture is inconsistent with $(1)$ and $(2)$ for $r=1$, hence it is also inconsistent with the OP's conjecture. Indeed, there exists an admissible $459$-tuple in $\{0,1,\dots,3240\}$, see here. Hence, by the $k$-tuple conjecture, there exists a $y\geq 2$ such that $\pi(3241+y)-\pi(y)\geq 459$. However, $(1)$ for $r=1$ and $x=3241$ would yield that $\pi(3241+y)-\pi(y)\leq\pi(3241)+1=458$, contradicting the previous inequality.

3. In fact the $k$-tuple conjecture is inconsistent with $(1)$ and $(2)$ for any integer $r\geq 0$. More precisely, assuming the $k$-tuple conjecture, Hensley and Richards proved in 1973 that for any $x\geq 2$ we have $$\sup_{y\geq x}\bigl(\pi(x+y)-\pi(x)-\pi(y)\bigr)\geq\bigl(\log 2-o(1)\bigr)\frac{x}{(\log x)^2}.$$