proof of Bernoulli’s inequality

Let I be the interval (-1,) and f:I the function defined as:


with α{0,1} fixed. Then f is differentiableMathworldPlanetmathPlanetmath and its derivativeMathworldPlanetmathPlanetmath is

f(x)=α(1+x)α-1-α, for all xI,

from which it follows that f(x)=0x=0.

  1. 1.

    If 0<α<1 then f(x)<0 for all x(0,) and f(x)>0 for all x(-1,0) which means that 0 is a global maximumMathworldPlanetmath point for f. Therefore f(x)<f(0) for all xI{0} which means that (1+x)α<1+αx for all x(-1,0).

  2. 2.

    If α[0,1] then f(x)>0 for all x(0,) and f(x)<0 for all x(-1,0) meaning that 0 is a global minimum point for f. This implies that f(x)>f(0) for all xI{0} which means that (1+x)α>1+αx for all x(-1,0).

Checking that the equality is satisfied for x=0 or for α{0,1} ends the proof.

Title proof of Bernoulli’s inequalityMathworldPlanetmath
Canonical name ProofOfBernoullisInequality
Date of creation 2013-03-22 12:38:14
Last modified on 2013-03-22 12:38:14
Owner danielm (240)
Last modified by danielm (240)
Numerical id 6
Author danielm (240)
Entry type Proof
Classification msc 26D99