# Bernoulli's Inequality/Proof 1

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## Contents

## Theorem

Let $x \in \R$ be a real number such that $x > -1$.

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

Then:

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

## Proof

Proof by induction:

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

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

### Basis for the Induction

$\map P 0$ is the case:

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

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:

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

We need to show that:

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

### Induction Step

This is our induction step:

\(\displaystyle \paren {1 + x}^{k + 1}\) | \(=\) | \(\displaystyle \paren {1 + x}^k \paren {1 + x}\) | |||||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle \paren {1 + k x} \paren {1 + x}\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 + \paren {k + 1} x + k x^2\) | |||||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle 1 + \paren {k + 1} x\) |

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

$\blacksquare$

## Source of Name

This entry was named for Jacob Bernoulli.