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# Prove a generalization of Bernoulli’s inequality

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 