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+0xis 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)

:$$egin{matrix}& iff & (1+x)^{k+1} ge 1+kx+x+kx^2 \& iff & (1+x)^{k+1} ge 1+(k+1)x+kx^2end{matrix}.

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

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External links

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