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 .

* Proof. *

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