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

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

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

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

Since $\alpha$ is in $J$, it is clear that if $a<0$, then

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

and, accordingly, if $a>0$ then

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

Thus, in either case, from 1 we deduce that

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

if $a<0$ 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:

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

This completes the proof.

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

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

. This can be proved using exactly the same method, by fixing $\alpha$ in the proof above in $\left(-1,0\right)$.

## Mathematics Subject Classification

26D99*no label found*

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