Bernoulli's inequality

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An illustration of Bernoulli's inequality, with the graphs of y=(1+x)^{r} and y=1+rx shown in red and blue respectively. Here, r=3.

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

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

(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, as shown below.

Proof of the inequality

For r = 0,

(1+x)^{0}\geq 1+0x\,

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

Now suppose the statement is true for r = k:

(1+x)^{k}\geq 1+kx.\,

Then it follows that

{\begin{aligned}&{}\qquad (1+x)(1+x)^{k}\geq (1+x)(1+kx)\quad {\text{(by hypothesis, since }}(1+x)\geq 0)\\&\iff (1+x)^{{k+1}}\geq 1+kx+x+kx^{2},\\&\iff (1+x)^{{k+1}}\geq 1+(k+1)x+kx^{2}.\end{aligned}}

However, as 1 + (k + 1)x + kx2 ≥ 1 + (k + 1)x (since kx2 ≥ 0), it follows that (1 + x)k + 1 ≥ 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 r ≥ 0.

Generalization

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.

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}\leq e^{{rx}},\!

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\geq 1 and 0\leq x\leq 1 is:

(1-x)^{t}\geq 1-xt.

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

t=1+1+\dots +1\geq 1+y+y^{2}+\ldots +y^{{t-1}}={\frac {1-y^{t}}{1-y}}

or equivalently xt\geq 1-(1-x)^{t}.

Proof for rational case

An "elementary" proof can be given using the fact that geometric mean of positive numbers is less than arithmetic mean

First assume t={\frac {a}{b}}\leq 1

By comparing Arithmetic and Geometric mean of b numbers ((1+x) occurs a times):

1,1,\ldots ,(1+x),(1+x),\ldots (1+x)

we get

(1+x)^{{a/b}}\leq \left(1+{\frac {a}{b}}x\right)

or equivalently

(1+x)^{{t}}\leq \left(1+tx\right)

This proves inequality for t\leq 1 case.

For s\geq 1 case, let z={\frac {a}{b}}x As x\geq -1,z\geq -{\frac {a}{b}} we get with s={\frac {1}{t}}={\frac {b}{a}}\geq 1,

\left(1+z\right)^{{s}}\geq 1+sz

This proves inequality for s\geq 1 case.

As these inequalities are true for all rational numbers t\leq 1 and s\geq 1, they are also true for all real numbers. this is because, any real number can be approximated by rational numbers to arbitrary precision (this formally follows from the Cauchy construction of real numbers).

References

  • Carothers, N. (2000). Real Analysis. Cambridge: Cambridge University Press. p. 9. ISBN 978-0-521-49756-5. 
  • Bullen, P.S. (1987). Handbook of Means and Their Inequalities. Berlin: Springer. p. 4. ISBN 978-1-4020-1522-9. 
  • Zaidman, Samuel (1997). Advanced Calculus. City: World Scientific Publishing Company. p. 32. ISBN 978-981-02-2704-3. 

External links

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