Generalization of exponent

The exponent ${\displaystyle r}$ can be generalized to an arbitrary real number as follows: if ${\displaystyle x>-1}$, then

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

for ${\displaystyle r\leq 0}$ or ${\displaystyle \geq 1}$, and

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

for ${\displaystyle 0\leq r\leq 1}$.

This generalization can be proved by comparing derivatives. The strict versions of these inequalities require ${\displaystyle x\neq 0}$ and ${\displaystyle r\neq 0,1}$.

Generalization of base

Instead of ${\displaystyle (1+x)^{n}}$ the inequality holds also in the form ${\displaystyle (1+x_{1})(1+x_{2})\dots (1+x_{r})\geq 1+x_{1}+x_{2}+\dots +x_{r}}$ where ${\displaystyle x_{1},x_{2},\dots ,x_{r}}$ are real numbers, all greater than ${\displaystyle -1}$, all with the same sign. Bernoulli's inequality is a special case when ${\displaystyle x_{1}=x_{2}=\dots =x_{r}=x}$. This generalized inequality can be proved by mathematical induction.

Proof

In the first step we take ${\displaystyle n=1}$. In this case the inequality ${\displaystyle 1+x_{1}\geq 1+x_{1}}$ is obviously true. In the second step we assume validity of the inequality for ${\displaystyle r}$ numbers and deduce validity for ${\displaystyle r+1}$ numbers. We assume that $${\displaystyle (1+x_{1})(1+x_{2})\dots (1+x_{r})\geq 1+x_{1}+x_{2}+\dots +x_{r}}$$ is valid. After multiplying both sides with a positive number ${\displaystyle (x_{r+1}+1)}$ we get: ${\displaystyle {\begin{alignedat}{2}(1+x_{1})(1+x_{2})\dots (1+x_{r})(1+x_{r+1})\geq &(1+x_{1}+x_{2}+\dots +x_{r})(1+x_{r+1})\\\geq &(1+x_{1}+x_{2}+\dots +x_{r})\cdot 1+(1+x_{1}+x_{2}+\dots +x_{r})\cdot x_{r+1}\\\geq &(1+x_{1}+x_{2}+\dots +x_{r})+x_{r+1}+x_{1}x_{r+1}+x_{2}x_{r+1}+\dots +x_{r}x_{r+1}\\\end{alignedat}}}$ As ${\displaystyle x_{1},x_{2},\dots x_{r},x_{r+1}}$ all have the same sign, the products ${\displaystyle x_{1}x_{r+1},x_{2}x_{r+1},\dots x_{r}x_{r+1}}$ are all positive numbers. So the quantity on the right-hand side can be bounded as follows: $${\displaystyle (1+x_{1}+x_{2}+\dots +x_{r})+x_{r+1}+x_{1}x_{r+1}+x_{2}x_{r+1}+\dots +x_{r}x_{r+1}\geq 1+x_{1}+x_{2}+\dots +x_{r}+x_{r+1},}$$ what was to be shown.

Arithmetic and geometric means

An elementary proof for ${\displaystyle 0\leq r\leq 1}$ and x ≥ -1 can be given using weighted AM-GM.

Let ${\displaystyle \lambda _{1},\lambda _{2}}$ be two non-negative real constants. By weighted AM-GM on ${\displaystyle 1,1+x}$ with weights ${\displaystyle \lambda _{1},\lambda _{2}}$ respectively, we get

${\displaystyle {\dfrac {\lambda _{1}\cdot 1+\lambda _{2}\cdot (1+x)}{\lambda _{1}+\lambda _{2}}}\geq {\sqrt[{\lambda _{1}+\lambda _{2}}]{(1+x)^{\lambda _{2}}}}.}$

Note that

${\displaystyle {\dfrac {\lambda _{1}\cdot 1+\lambda _{2}\cdot (1+x)}{\lambda _{1}+\lambda _{2}}}={\dfrac {\lambda _{1}+\lambda _{2}+\lambda _{2}x}{\lambda _{1}+\lambda _{2}}}=1+{\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}x}$

and

${\displaystyle {\sqrt[{\lambda _{1}+\lambda _{2}}]{(1+x)^{\lambda _{2}}}}=(1+x)^{\frac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}},}$

so our inequality is equivalent to

${\displaystyle 1+{\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}x\geq (1+x)^{\frac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}.}$

After substituting ${\displaystyle r={\dfrac {\lambda _{2}}{\lambda _{1}+\lambda _{2}}}}$ (bearing in mind that this implies ${\displaystyle 0\leq r\leq 1}$) our inequality turns into

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

which is Bernoulli's inequality.

Geometric series

Bernoulli's inequality

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

(1)

is equivalent to

${\displaystyle (1+x)^{r}-1-rx\geq 0,}$

(2)

and by the formula for geometric series (using y = 1 + x) we get

${\displaystyle (1+x)^{r}-1=y^{r}-1=\left(\sum _{k=0}^{r-1}y^{k}\right)\cdot (y-1)=\left(\sum _{k=0}^{r-1}(1+x)^{k}\right)\cdot x}$

(3)

which leads to

${\displaystyle (1+x)^{r}-1-rx=\left(\left(\sum _{k=0}^{r-1}(1+x)^{k}\right)-r\right)\cdot x=\left(\sum _{k=0}^{r-1}\left((1+x)^{k}-1\right)\right)\cdot x\geq 0.}$

(4)

Now if ${\displaystyle x\geq 0}$ then by monotony of the powers each summand ${\displaystyle (1+x)^{k}-1=(1+x)^{k}-1^{k}\geq 0}$, and therefore their sum is greater ${\displaystyle 0}$ and hence the product on the LHS of (4).

If ${\displaystyle 0\geq x\geq -2}$ then by the same arguments ${\displaystyle 1\geq (1+x)^{k}}$ and thus all addends ${\displaystyle (1+x)^{k}-1}$ are non-positive and hence so is their sum. Since the product of two non-positive numbers is non-negative, we get again (4).

Binomial theorem

One can prove Bernoulli's inequality for x ≥ 0 using the binomial theorem. It is true trivially for r = 0, so suppose r is a positive integer. Then ${\displaystyle (1+x)^{r}=1+rx+{\tbinom {r}{2}}x^{2}+...+{\tbinom {r}{r}}x^{r}.}$ Clearly ${\displaystyle {\tbinom {r}{2}}x^{2}+...+{\tbinom {r}{r}}x^{r}\geq 0,}$ and hence ${\displaystyle (1+x)^{r}\geq 1+rx}$ as required.

Using convexity

For ${\displaystyle 0\neq x\geq -1}$ the function ${\displaystyle h(\alpha )=(1+x)^{\alpha }}$ is strictly convex. Therefore for ${\displaystyle 0<\alpha <1}$ holds

${\displaystyle (1+x)^{\alpha }=h(\alpha )=h((1-\alpha )\cdot 0+\alpha \cdot 1)<(1-\alpha )h(0)+\alpha h(1)=1+\alpha x}$

and the reversed inequality is valid for ${\displaystyle \alpha <0}$ and ${\displaystyle \alpha >1}$.