For real numbers with for all and all of the having the same sign, prove
As a special case let and prove Bernoulli’s inequality,
Finally, show that if then equality holds only when .
Proof. The proof is by induction. For the case , we have,
so the inequality holds for .
Assume then that the inequality holds for some . Then,
But, since every must have the same sign (thus, and must have the same sign, so the product is positive). Thus,
Hence, the inequality holds for the case ; and therefore, for all
Now, if where and we apply the theorem above to obtain Bernoulli’s inequality,
Claim: Equality holds in Bernoulli’s inequality if and only if .
If then , so indeed equality holds for . Next, we use induction to show that if , then the inequality must be strict. (Hence, equality holds if and only if .)
For the case , on the left we have,
since for . So, the inequality is strict for the case . Assume then that the inequality is strict for some . Then,
Where the final line follows since and implies . Therefore, the inequality is strict for all if .
Hence, the equality holds if and only if