To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)
Top

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.

Related Calculators
Calculator for Inequalities Inequality Calculator
Compound Inequalities Calculator Graph Inequalities Calculator
 

Definition

Back to Top
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

Back to Top
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

Back to Top
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

Back to Top
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

Back to Top
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 $y\ =\ 1\ -\ x$

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

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

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

$\Rightarrow$ $r\ x$ = $1\ –\ (1\ –\ x)^r$
Related Topics
Math Help Online Online Math Tutor
*AP and SAT are registered trademarks of the College Board.