Doing it with the help of induction on the number of elements in A.
Suppose Statement P(n) : If n be the number of elements in A and B be any non empty subset in Z then |A + B| >= |A| + |B| -1
Basis Step: To prove P(1) is true
P(1) : If there is only 1 element in A then |A + B| >= |A| + |B| -1
If there is 1 element in A then |A+B| = |B| >= |A| + |B| -1 = |B|
Hence P(1) is true.
Inductive Hypothesis : We assume P(k) is true. That is If k be the number of elements in A then |A + B| >= |A| + |B| -1
Inductive Step : With the help of inductive hypothesis we prove that P(k+1) is true.
So, now |A| = k+1.
Now, since A is finite so A must contain a least element ai.
Now, |A\ai| = k .
So, from inductive hypothesis we get
|A\ai + B| >= |A\ai| + |B| -1
=> |A\ai + B| +1 >= |A\ai| + |B| -1 +1 = |A| + |B| -1....................(1)
Now, since is finite it must contain a least element say bi.
Note that the element (ai + bi) will belong to (A + B) but will not belong to (A\ai + B)
So, |(A + B)| >= |(A\ai + B)| + 1 ......................(2)
So, from the inequalities 1 and 2 we get
|A + B| >= |A| + |B| -1
Hence P(k+1) is true.
Thus whenever P(k) is true we will have P(k+1) is true.
Thus P(n) is true for all n belongs to Z.
Hence |A + B| >= |A| + |B| -1 is true for all non empty finite subsets of Z.