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 $r\in J=(0,\mathrm{\infty })$. Observethat if $x=0$ the inequality holds quite obviously. Let us nowconsider the case where $x\ne 0$. Consider now thefunction $f:I\text{x}J\to ℝ$ 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,

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

Considertwo points $a\ne 0$ in $I$ and $0$ in $I$. Then clearly by themean value theorem, for any arbitrary, fixed $\alpha$ in $J$, thereexists 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 weconclude that, in either case,${(1+a)}^{\alpha }-1-\alpha a>0$. Thatis,

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

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 $J$ wehave:

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

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 $(-1,0)$.