Bernoulli's inequality

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 ≥ −2. 
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.

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

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".

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 suppose the 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 integer r.

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 (1 + 1/k)k < e.

Alternative form
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

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  (bearing in mind that this implies ) 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 (using y=1+x) 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

(1) For , 
Obviously, 

Thus, 

(2) For , 

(3) 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 by a 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.

Notes

References

External links 
 
 Bernoulli Inequality by Chris Boucher, Wolfram Demonstrations Project.
 

Category:Inequalities