# 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:

$\paren {1 - x}^n \ge 1 - n x$

### General Result

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

$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$

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:

$\paren {1 + y}^n \ge 1 + n y$

Thus:

$\paren {1 + \paren {-x} }^n \ge 1 + n \paren {-x}$

Hence the result.

$\blacksquare$

## Proof 2

Proof by induction:

Let $0 < x < 1$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\paren {1 - x}^n \ge 1 - n x$

### Basis for the Induction

$\map P 0$ is the case:

$\paren {1 - x}^0 = 1 \ge 1 - 0 x = 1$

so $\map P 0$ holds.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

$\paren {1 - x}^k \ge 1 - k x$

We need to show that:

$\paren {1 - x}^{k + 1} \ge 1 - \paren {k + 1} x$

### Induction Step

This is our induction step:

 $\ds \paren {1 - x}^{k + 1}$ $=$ $\ds \paren {1 - x}^k \paren {1 - x}$ $\ds$ $\ge$ $\ds \paren {1 - k x} \paren {1 - x}$ Induction Hypothesis $\ds$ $=$ $\ds 1 - \paren {k + 1} x + k x^2$ $\ds$ $\ge$ $\ds 1 - \paren {k + 1} x$

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

$\blacksquare$