Bernoulli's inequality

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An illustration of Bernoulli's inequality, with the graphs of and shown in red and blue respectively. Here,

In real analysis, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x.

The inequality states that

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

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.


Jacob Bernoulli first published the inequality in his treatise “Positiones Arithmeticae de Seriebus Infinitis” (Basel, 1689), where he used the inequality often.[1]

According to Joseph E. Hofmann, Über die Exercitatio Geometrica des M. A. Ricci (1963), p. 177, the inequality is actually due to Sluse in his Mesolabum (1668 edition), Chapter IV "De maximis & minimis".[1]

Proof of the inequality[edit]

For r = 0,

is equivalent to 1 ≥ 1 which is true as required.

Now suppose the statement is true for r = k:

Then it follows that

By induction we conclude the statement is true for all r ≥ 0.


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[edit]

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

where e = 2.718.... This may be proved using the inequality (1 + 1/k)k < e.

Alternative form[edit]

An alternative form of Bernoulli's inequality for and is:

This can be proved (for integer t) by using the formula for geometric series: (using y=1-x)

or equivalently

Alternative Proof[edit]

Using AM-GM

An elementary proof for 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


so our inequality is equivalent to

After substituting (bearing in mind that this implies ) our inequality turns into

which is Bernoulli's inequality.

Using Binomial theorem

(1) For x > 0, Obviously,


(2) For x = 0, it is obvious that

(3) For −1 ≤ x < 0, let y = −x, then 0 < y ≤ 1

Replace x with −y, we have

Also, according to the binomial theorem,


Notice that

Therefore, we can see that each binomial term is multiplied by a factor , and that will make each term smaller than the term before.

For that reason,


Replace y with −x back, we get

Notice that by using binomial theorem, we can only prove the cases when r is a positive integer or zero.



  • Carothers, N.L. (2000). Real analysis. Cambridge: Cambridge University Press. p. 9. ISBN 978-0-521-49756-5. 
  • Bullen, P. S. (2003). Handbook of means and their inequalities. Dordercht [u.a.]: Kluwer Academic Publ. p. 4. ISBN 978-1-4020-1522-9. 
  • Zaidman, S. (1997). Advanced calculus : an introduction to mathematical analysis. River Edge, NJ: World Scientific. p. 32. ISBN 978-981-02-2704-3. 

External links[edit]