proof of Bernoulli’s inequality employing the mean value theorem

Let us take as our assumptionPlanetmathPlanetmath that xI=(-1,) and that rJ=(0,). Observe that if x=0 the inequalityMathworldPlanetmath holds quite obviously. Let us now consider the case where x0. Consider now the function f:IxJ given by


Observe that for all r in J fixed, f is, indeed, differentiableMathworldPlanetmathPlanetmath on I. In particular,


Consider two points a0 in I and 0 in I. Then clearly by the mean value theorem, for any arbitrary, fixed α in J, there exists a c in I such that,

fxf(c,α)=(1+a)α-1-αaa (1)

Since α is in J, it is clear that if a<0, then


and, accordingly, if a>0 then


Thus, in either case, from 1 we deduce that


if a<0 and


if a>0. From this we conclude that, in either case,(1+a)α-1-αa>0. That is,


for all choices of a in I-{0} and all choices of α in J. If a=0 in I, we have


for all choices of α in J. Generally, for all x in I and all r in J we have:


This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Notice that if r is in (-1,0) then the inequality would be reversed. That is:


. This can be proved using exactly the same method, by fixing α in the proof above in (-1,0).

Title proof of Bernoulli’s inequality employing the mean value theorem
Canonical name ProofOfBernoullisInequalityEmployingTheMeanValueTheorem
Date of creation 2013-03-22 15:49:53
Last modified on 2013-03-22 15:49:53
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 10
Author rspuzio (6075)
Entry type Proof
Classification msc 26D99