I am taking this argument from Gelfond & Linnik's Elementary Methods in the Analytic Theory of Numbers.
They state if for every $n \ge 1$, there exists $m \in [1,n]$ where $C(n) - C(n-m) \ge (d(A) + d(B))m$, then it "may be deduced from it without difficulty" that $d(C) \ge d(A) + d(B)$
Here's the reasoning presented:
(1) Assume that for every $n \ge 1$, there exists $m \in [1,n]$ such that $C(n) - C(n-m) \ge (d(A) + d(B))m$
(2) Consider the segment $[1,n-m]$
(3) Then, it follows that a segment $[n - m - m' + 1, n - m]$ can be found where $C(n-m) - C(n-m-m') \ge (d(A)+d(B))m'$
(4) We can extract from $[1,n-m-m']$ the "extreme right-hand segment with the same property then continue our argument."
(5) "As a result we obviously arrive at the inequality": $d(C) \ge d(A) + d(B)$
This might be obvious to the others. I'm still scratching my head. If we can make the size of $m$ as small as we want, why would this prove that $d(C) \ge d(A) + d(B)$.
It would help me to understand the argument if someone could explain the principle that lets us reach the conclusion intended.
Here's my understanding of Schnirelmann Density.
$A,B$ are infinite sequences of integers starting with $0$ with in sequential order such as $0, a_1, a_2, \cdots$ where $0 < a_1 < a_2 < \cdots$
Schnirelmann density is defined as: $$d(A) = \inf\limits_{n}\frac{A(n)}{n}$$
where: $$A(n) = \sum\limits_{0<a_i\le{n}}{1}$$
So, it is clear that: $$0 \le \frac{A(n)}{n} \le 1$$
$C = A+B$ where $+$ is the sumset.
As we repeat the argument on ever-smaller segments, we get:
$C(n) - C(n-m) \geq (d(A) + d(B))m$
$C(n-m) - C(n-m-m^\prime) \geq (d(A)+d(B))m^\prime$
$C(n-m-m^\prime) - C(n-m-m^\prime-m^{\prime\prime}) \geq (d(A)+d(B))m^{\prime\prime}$
and so on. We add these and get the LHS to telescope, so
$C(n) \geq (d(A)+d(B))(m+m^\prime+m^{\prime\prime}+\cdots) = (d(A)+d(B))n$.
Now divide both sides by $n$: $\displaystyle \frac{C(n)}{n} \geq d(A)+d(B)$. Hence taking the infimum over $n$ the inequality still holds and $d(C) \geq d(A)+d(B)$.