In real analysis, Bernoulli's inequality is an inequality that approximates exponentiations of 1 + x. The inequality states that for every integerr ≥ 0 and every real numberx ≥ −1. If the exponentr is even, then the inequality is valid for allreal numbersx. The strict version of the inequality reads for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0. There is also a generalized version that says for every real number r ≥ 1 and real number x ≥ -1, while for 0 ≤ r ≤ 1 and real number x ≥ -1, Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction, as shown below.
History
Jacob Bernoulli first published the inequality in his treatise “Positiones Arithmeticae de Seriebus Infinitis”, where he used the inequality often. According to Joseph E. Hofmann, Über die Exercitatio Geometrica des M. A. Ricci, p. 177, the inequality is actually due to Sluse in his Mesolabum, Chapter IV "De maximis & minimis".
Proof of the inequality
We proceed with mathematical induction in the following form:
we prove the inequality for,
from validity for some r we deduce validity for r+2.
For r = 0, is equivalent to 1 ≥ 1 which is true as required. Similarly, for r = 1 we have Now supposethe statement is true for r = k: Then it follows that since as well as. By the modified induction we conclude the statement is true for every non-negative integerr.
Generalization
The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then for r ≤ 0 or r ≥ 1, and for 0 ≤ r ≤ 1. This generalization can be proved by comparing derivatives. Again, the strict versions of these inequalities require x ≠ 0 and r ≠ 0, 1.
Related inequalities
The following inequality estimates the r-th power of 1 + x from the other side. For any real numbers x, r with r > 0, one has where e = 2.718.... This may be proved using the inequality k < e.
Alternative form
An alternative form of Bernoulli's inequality for and is: This can be proved by using the formula for geometric series: or equivalently
Alternative Proof
Using AM-GM An elementary proof for and x ≥ -1 can be given using Weighted AM-GM. Let be two non-negative real constants. By Weighted AM-GM on with weights respectively, we get Note that and so our inequality is equivalent to After substituting our inequality turns into which is Bernoulli's inequality. Using the formula for geometric series Bernoulli's inequality is equivalent to and by the formula for geometric series we get
which leads to Now if then by monotony of the powers each summand, therefore their sum is greater and hence the product on the LHS of. If then by the same arguments and thus all addends are non-positive and hence their sum. Since the product of two non-positive numbers is non-negative, we get again . Using Binomial theorem For, Obviously, Thus, For, For, let, then Replace with, we have Also, according to the binomial theorem, then Notice that Therefore, we can see that each binomial term is multiplied bya factor , and that will make each term smaller than the term before. For that reason, Hence, Replace with back, we get Notice that by using binomial theorem, we can only prove the cases when r is a positive integer or zero.
