Given . Then if
prove that
with equality if and only if .
Proof. We consider two cases: the first one is the case that




Case #1: If

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
Thus,
Hence, the inequality holds for ; and therefore, for all