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Encyclopedia > Bernoulli's inequality

In real analysis, Bernoulli's inequality is an inequality that approximates exponentiations of 1 + x. Real analysis is a branch of mathematical analysis dealing with the set of real numbers and functions of real numbers. ... The feasible regions of linear programming are defined by a set of inequalities. ... Exponentiation is a mathematical operation, written an, involving two numbers, the base a and the exponent n. ...


The inequality states that

(1 + x)^r geq 1 + rx!

for every integer r ≥ 0 and every real number x > −1. If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads The integers are commonly denoted by the above symbol. ... In mathematics, the real numbers may be described informally as numbers that can be given by an infinite decimal representation, such as 2. ... In mathematics, any integer (whole number) is either even or odd. ...

(1 + x)^r > 1 + rx!

for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0.


Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction: In mathematics, a proof is a demonstration that, given certain axioms, some statement of interest is necessarily true. ... Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. ...



Proof: For r=0, (1+x)^0 ge 1+0x equals 1ge 1 which is true as required.


Now suppose the statement is true for r=k: (1+x)^k ge 1+kx Then it follows that (1+x)^{k+1} = (1+x)(1+x)^k ge (1+x)(1+kx) (by hypothesis, since (1+x)ge 0) = 1+kx+x+kx^2 = 1+(k+1)x + kx^2 ge 1+(k+1)x (since kx^2 ge 0) So it follows that (1+x)^{k+1} ge 1+(k+1)x, which means the statement is true for r=k+1 as required.


By induction we conclude the statement is true for all rge 0   Box ;



The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then

(1 + x)^r geq 1 + rx!

for r ≤ 0 or r ≥ 1, and

(1 + x)^r leq 1 + rx!

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. In some places this article assumes an acquaintance with algebra, analytic geometry, or the limit. ...


Related inequalities

The following inequality estimates the r-th power of 1 + x from the other side. For any real numbers x, r > 0, one has

(1 + x)^r < e^{rx},!

where e = 2.718.... This may be proved using the inequality (1 + 1/k)k < e. The title given to this article is incorrect due to technical limitations. ...



 

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