Given . Then if
with equality if and only if .
Proof. We consider two cases: the first one is the case that , and the second is the case that for at least one in .
Case #1: If , then
Thus, the inequality holds (in fact, equality holds).
Case #2: Now for the case that at least one of the . The proof is by induction on the number of terms . For the case , we have , and by assumption at least one of these is not equal to 1. But, if one of them is not 1, then neither is the other (since the product must equal 1). So,
Where the final inequality follows since since by assumption. Therefore, the inequality indeed holds for the case .
Assume then that the inequality holds for some . Then, let with for at least one . If , then there must be some such that (otherwise, if for all , then ). Similarly, if , then there is some such that . Hence, we have a pair, with one member of the pair greater than 1, and the other less than 1. Without loss of generality (since multiplication and addition are commutative), let this pair be and . Then, define . So,
Further, since (since one of is greater than 1 and the other is less than 1, so one of is positive and the other is negative), we have
Hence, the inequality holds for ; and therefore, for all