# Bernoulli's Inequality/Proof 1

## 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.