# Bernoulli's Inequality/Corollary

## Theorem

Let $x \in \R$ be a real number such that $0 < x < 1$.

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

$\left({1 - x}\right)^n \ge 1 - n x$

### General Result

For all $n \in \Z_{\ge 0}$:

$\displaystyle \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$

where $0 < a_j < 1$ for all $j$.

## Proof 1

Let $0 < x < 1$.

Let $y = -x$.

Then $y > -1$ and by Bernoulli's Inequality:

$\left({1 + y}\right)^n \ge 1 + n y$

Thus:

$\left({1 + \left({-x}\right)}\right)^n \ge 1 + n \left({-x}\right)$

Hence the result.

$\blacksquare$

## Proof 2

Proof by induction:

Let $0 < x < 1$.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\left({1 - x}\right)^n \ge 1 - n x$

### Basis for the Induction

$P \left({0}\right)$ is the case:

$\left({1 - x}\right)^0 = 1 \ge 1 - 0 x = 1$

so $P \left({0}\right)$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\left({1 - x}\right)^k \ge 1 - k x$

We need to show that:

$\left({1 - x}\right)^{k+1} \ge 1 - \left({k + 1}\right) x$

### Induction Step

This is our induction step:

 $\displaystyle \left({1 - x}\right)^{k + 1}$ $=$ $\displaystyle \left({1 - x}\right)^k \left({1 - x}\right)$ $\displaystyle$ $\ge$ $\displaystyle \left({1 - k x}\right) \left({1 - x}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle 1 - \left({k + 1}\right) x + k x^2$ $\displaystyle$ $\ge$ $\displaystyle 1 - \left({k + 1}\right) x$

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$