Bernoulli's inequality

In real analysis, Bernoulli's inequality 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 $r=0,,$:$(1+x)^0\; ge\; 1+0x$is equivalent to $1ge\; 1$ which is true as required.

Now suppose the statement is true for $r=k$::$(1+x)^k\; ge\; 1+kx.$Then it follows that:$(1+x)(1+x)^k\; ge\; (1+x)(1+kx)$ (by hypothesis, since $(1+x)ge\; 0$)

:

However, as $1+(k+1)x\; +\; kx^2\; ge\; 1+(k+1)x$ (since $kx^2\; ge\; 0$), 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 $rge\; 0.$

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" &gt; 0, one has:$(1\; +\; x)^r\; le\; e^\{rx\},!$where "e" = 2.718....This may be proved using the inequality (1 + 1/"k")^"k" < "e".

References

*
*
*

External links

*
* [http://demonstrations.wolfram.com/BernoulliInequality/ Bernoulli Inequality] by Chris Boucher, The Wolfram Demonstrations Project.