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# Bernoulli's Inequality

Bernoulli’s inequality is applicable on real numbers. It was named after Jacob Bernoulli. This inequality is used in approximation process of exponentiations of form $1\ +\ x$. We will discuss it only on integer cases here.

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## Definition

According to Bernoulli’s inequality, if ‘r’ is an integer such that $r\ \geq\ 0$, and x is a real number such that $x\ \geq\ -1$, then
$(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$
For the strict version of this inequality r must be greater than or equal to $2$, that is $r\ \geq\ 2$ for all real numbers $x$ greater than or equal to $-1$ but not equal to zero, that is $x\ \geq\ -1$ and $x \neq 0$. Then the inequality becomes

$(1\ +\ x)^r\ >\ 1\ +\ r\ x$

## Proof of Bernoulli’s Inequality

We consider the arithmetic and geometric mean of $r$ positive numbers $(1\ +\ r\ x)$, $1$, $1$, … , $1$ where we have $(r\ –\ 1)$ times $1$.

We know that arithmetic mean $\geq$ geometric mean

$\Rightarrow$ $\frac{[(1\ +\ r\ x)\ +\ 1\ +\ 1\ ….\ +\ 1]}{r}$ $\geq$ rth root $((1\ +\ r\ x)\ \times 1 \times 1 \times … \times 1)$

$\Rightarrow$ $\frac{(r\ +\ r\ x)}{r}$ $\geq$ rth root $(1\ +\ r\ x)$

$\Rightarrow$ $1\ +\ x\ \geq$ rth root $(1\ +\ r\ x)$

$\Rightarrow$ $(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$

We can also prove it using Binomial Theorem

If $x\ >\ 0$, then

$(1\ +\ x)^{r}$ = $1\ +\ r \ x\ +\ ^{r}\textrm{C}_2 x^2\ +\ ^{r}\textrm{C}_3 x^3 +\ ^{r}\textrm{C}_4 x^4 +\ ^{r}\textrm{C}_5 x^5 +\ …\ +\ ^{r}\textrm{C}_r x^r$

$\Rightarrow$ $(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$

If $x\ =\ 0$, then

$(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$ for obvious reasons

If $-1\ <\ x\ <\ 0$, then put $y\ =\ -x$, we get

$0\ <\ y\ <\ 1$ and $(1\ –\ y)^{r}$ = $1\ –\ r\ y$ + $^{r}\textrm{C}_2 y^{2}$ – $^{r}\textrm{C}_3 y^{3}$ + $^{r}\textrm{C}_4 y^{4}$ $^{r}\textrm{C}_5 y^{5} +\ …\ +\ (-1)^{r}$ $^{r}\textrm{C}_r y^r$

When $y$ = $1$, we get

$(1\ –\ 1)^r$ = $1\ –\ r\ +\ ^{r}\textrm{C}_2 –\ ^{r}\textrm{C}_3 +\ ^{r}\textrm{C}_4 –\ ^{r}\textrm{C}_5 +\ …\ +\ (-1)^r\ ^{r}\textrm{C}_r$

$\Rightarrow$ $^{r}\textrm{C}_2 –\ ^{r}\textrm{C}_3 +\ ^{r}\textrm{C}_4 –\ ^{r}\textrm{C}_5 +\ …\ +\ (-1)^r\ ^{r}\textrm{C}_r$ = $r\ –\ 1$

$\Rightarrow$ $^{r}\textrm{C}_2 –\ ^{r}\textrm{C}_3 +\ ^{r}\textrm{C}_4 –\ ^{r}\textrm{C}_5 +\ …\ +\ (-1)^r\ ^{r}\textrm{C}_r >\ 0$

When $0\ <\ y\ <\ 1,\ y^2\ >\ y^3\ >\ …\ >\ y^r$

Then, when we multiply above with their y terms we get

$^{r}\textrm{C}_2 y^2 –\ ^{r}\textrm{C}_3\ y^3\ +\ ^{r}\textrm{C}_4\ y^4\ –\ ^{r}\textrm{C}_5\ y^5\ +\ …\ +\ (-1)^r\ ^{r}\textrm{C}_r\ y^r\ \geq\ 0$

$\Rightarrow$ $(1\ –\ y)^r\ \geq\ 1\ –\ r\ y$ for $0\ <\ y\ <\ 1$

$\Rightarrow$ $(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$ for $-1\ <\ x\ <\ 0$

## Bernoulli’s Inequality Induction Proof

We can also Bernoulli’s inequality using the principle of mathematical induction.

According to the inequality we have $(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$

When $r\ =\ 0$,

$(1\ +\ x)^0\ \geq\ 1\ +\ 0$

$\Rightarrow$ $1\ \geq\ 1$, which is always true.

Let the statement hold true for some $r\ =\ j$.

That is $(1\ +\ x)^j\ \geq\ 1\ +\ j\ x$

Now consider, $(1\ +\ x)^{(j\ +\ 1)}$

$(1\ +\ x)^{(j\ +\ 1)}$ = $(1\ +\ x)^j\ \times (1\ +\ x)$

$\Rightarrow$ $(1\ +\ x)^{(j\ +\ 1)}\ \geq\ (1\ +\ j\ x)\ (1\ +\ x)$

$\Rightarrow$ $(1\ +\ x)^{(j\ +\ 1)}\ \geq\ 1\ +\ j\ x^2\ +\ (j\ +\ 1)\ x$

$\Rightarrow$ $(1\ +\ x)^{(j\ +\ 1)}\ \geq\ 1\ +\ (j\ +\ 1)\ x$

Thus, the statement holds true for $r\ =\ j\ +\ 1$. Hence by the principle of mathematical induction the inequality holds true for all $r\ \geq\ 0$.

## Extension of Bernoulli’s Inequality

When $x\ >\ -1$, then we have

1. $(1\ +\ x)^r\ \leq\ 1\ +\ r\ x$ for $0\ <\ r\ <\ 1$

2. $(1\ +\ x)^r\ \geq\ 1\ +\ r\ x$ for $r\ <\ 0$ and $r\ >\ 1$

This can be easily proven using the Bernoulli’s main inequality and then substituting $x\ >\ -1$ with different $r$ as given in interval.

This can also be proved easily by comparing the derivatives. For the strict versions in the inequalities we require to have $x$ not equal to zero and  $r$ not equal to $0$ and $1$,
that is $x \neq 0$, $r \neq 0$, $1$.

## Alternate Form

There is an alternative form of Bernoulli’s inequality for $r\ \geq\ 1$ and $0\ \leq\ x\ \leq\ 1$.

In this case, the inequality transforms to $(1\ –\ x)^r\ \geq\ 1\ –\ r\ x$

This can be easily proved by the use of the formula for geometric series for $r$.

Let $y\ =\ 1\ -\ x$

$r$ = $1\ +\ 1\ +\ 1\ +\ …\ +\ \geq\ 1\ +\ y\ +\ y^2\ +\ ….\ +\ y^{(r\ –\ 1)}$ = $\frac{(1\ –\ y^r)}{(1\ –\ y)}$

$\Rightarrow$ $r$ = $\frac{(1\ –\ (1\ –\ x)^r)}{(1 – (1 – x))}$

$\Rightarrow$ $r$ = $\frac{(1\ –\ (1\ –\ x)^r)}{x}$

$\Rightarrow$ $r\ x$ = $1\ –\ (1\ –\ x)^r$
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