Bernoulli's inequality - Alchetron, the free social encyclopedia

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

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

For r = 0,

( 1 + x ) 0 ≥ 1 + 0 x

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

Now suppose the statement is true for r = k:

( 1 + x ) k ≥ 1 + k x .

Then it follows that

( 1 + x ) ( 1 + x ) k ≥ ( 1 + x ) ( 1 + k x ) (by hypothesis, since ( 1 + x ) ≥ 0 ) ⟺ ( 1 + x ) k + 1 ≥ 1 + k x + x + k x 2 , ⟺ ( 1 + x ) k + 1 ≥ 1 + ( k + 1 ) x + k x 2 . ⟺ ( 1 + x ) k + 1 ≥ 1 + ( k + 1 ) x

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

Generalization

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

( 1 + x ) r ≥ 1 + r x

for r ≤ 0 or r ≥ 1, and

( 1 + x ) r ≤ 1 + r x

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.

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 r x ,

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 t ≥ 1 and 0 ≤ x ≤ 1 is:

( 1 − x ) t ≥ 1 − x t .

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

t = 1 + 1 + ⋯ + 1 ≥ 1 + y + y 2 + … + y t − 1 = 1 − y t 1 − y

or equivalently x t ≥ 1 − ( 1 − x ) t .

Alternative Proof

Using AM-GM

An elementary proof for 0 ≤ r ≤ 1 can be given using Weighted AM-GM.

Let λ 1 , λ 2 be two non-negative real constants. By Weighted AM-GM on 1 , 1 + x with weights λ 1 , λ 2 respectively, we get

λ 1 ⋅ 1 + λ 2 ⋅ ( 1 + x ) λ 1 + λ 2 ≥ ( 1 + x ) λ 2 λ 1 + λ 2

Note that

λ 1 ⋅ 1 + λ 2 ⋅ ( 1 + x ) λ 1 + λ 2 = λ 1 + λ 2 + λ 2 x λ 1 + λ 2 = 1 + λ 2 λ 1 + λ 2 x

and

( 1 + x ) λ 2 λ 1 + λ 2 = ( 1 + x ) λ 2 λ 1 + λ 2

so our inequality is equivalent to

1 + λ 2 λ 1 + λ 2 x ≥ ( 1 + x ) λ 2 λ 1 + λ 2

After substituting r = λ 2 λ 1 + λ 2 (bearing in mind that this implies 0 ≤ r ≤ 1 ) our inequality turns into

1 + r x ≥ ( 1 + x ) r which is Bernoulli's inequality.

Using Binomial theorem

(1) For x > 0, ( 1 + x ) r = 1 + r x + ( r 2 ) x 2 + . . . + ( r r ) x r Obviously, ( r 2 ) x 2 + . . . + ( r r ) x r ≥ 0

Thus, ( 1 + x ) r ≥ 1 + r x

(2) For x = 0, it is obvious that ( 1 + x ) r = 1 + r x

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

Replace x with −y, we have ( 1 − y ) r = 1 − r y + ( r 2 ) y 2 + . . . + ( − 1 ) r ( r r ) y r

Also, according to the binomial theorem, 0 = 1 − r + ( r 2 ) + . . . + ( − 1 ) r ( r r )

then ( r 2 ) + . . . + ( − 1 ) r ( r r ) = r − 1 ≥ 0

Notice that y 2 ≥ y 3 ≥ . . . ≥ y r

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

For that reason, ( r 2 ) y 2 + . . . + ( − 1 ) r ( r r ) y r ≥ 0

Hence, ( 1 − y ) r = 1 − r y + ( r 2 ) y 2 + . . . + ( − 1 ) r ( r r ) y r ≥ 1 − r y

Replace y with −x back, we get ( 1 + x ) r ≥ 1 + r x

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

References

Bernoulli's inequality Wikipedia

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