# proof of Bernoulli’s inequality employing the mean value theorem

Let us take as our assumption^{} that $x\in I=(-1,\mathrm{\infty})$ and that $r\in J=(0,\mathrm{\infty})$. Observe
that if $x=0$ the inequality^{} holds quite obviously. Let us now
consider the case where $x\ne 0$. Consider now the
function $f:I\text{x}J\to \mathbb{R}$ given by

$$f(x,r)={(1+x)}^{r}-1-rx$$ |

Observe that for all $r$ in $J$
fixed, $f$ is, indeed, differentiable^{} on $I$. In particular,

$$\frac{\partial}{\partial x}f(x,r)=r{(1+x)}^{r-1}-r$$ |

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

$${f}_{x}^{\prime}(c,\alpha )=\frac{f(a,\alpha )-f(0,\alpha )}{a}$$ |

$$\iff {f}_{x}^{\prime}f(c,\alpha )=\frac{{(1+a)}^{\alpha}-1-\alpha a}{a}$$ | (1) |

Since $\alpha $ is in $J$, it is clear that if $$, then

$$ |

and, accordingly, if $a>0$ then

$${f}_{x}^{\prime}(a,\alpha )>0$$ |

Thus, in either case, from 1 we deduce that

$$ |

if $$ and

$$\frac{{(1+a)}^{\alpha}-1-\alpha a}{a}>0$$ |

if $a>0$. From this we conclude that, in either case,${(1+a)}^{\alpha}-1-\alpha a>0$. That is,

$${(1+a)}^{\alpha}>1+\alpha a$$ |

for all choices of $a$ in $I-\left\{0\right\}$ and all choices of $\alpha $ in $J$. If $a=0$ in $I$, we have

$${(1+a)}^{\alpha}=1+\alpha a$$ |

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

$${(1+x)}^{r}\ge 1+rx$$ |

This completes^{} the proof.

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

$${(1+x)}^{r}\le 1+rx$$ |

. This can be proved using exactly the same method, by fixing $\alpha $ 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 |