# 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 $\left(-1,0\right)$.

Title proof of Bernoulli’s inequality employing the mean value theorem ProofOfBernoullisInequalityEmployingTheMeanValueTheorem 2013-03-22 15:49:53 2013-03-22 15:49:53 rspuzio (6075) rspuzio (6075) 10 rspuzio (6075) Proof msc 26D99