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



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> Prove: $$P\_{n+m} \ge P\_n + P\_m .$$
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Can you give a hint, reference, comment, or proof?



Prove: If $P_n$ is $n$-$th$ prime number then $P_{n+m} \ge P_n+P_m$
This is an expanded version of my previous answer. It shows, among other things, that the OP's conjecture contradicts the [$k$-tuple conjecture](http://mathworld.wolfram.com/k-TupleConjecture.html) 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](http://math.mit.edu/%7Eprimegaps/tuples/admissible_459_3240.txt). 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](http://matwbn.icm.edu.pl/ksiazki/aa/aa25/aa2548.pdf) proved in 1973 that for any $x\geq 2$ we have
$$\sup\_{y\geq x}\bigl(\pi(x+y)-\pi(x)-\pi(y)\bigr)\geq\bigl(\log 2-o(1)\bigr)\frac{x}{(\log x)^2}.$$


