Bernoulli's Inequality

Theorem

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

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

Then:

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

Corollary

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

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

Then:

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

Proof 1

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:

\(\ds \paren {1 + x}^{k + 1}\)

\(=\)

\(\ds \paren {1 + x}^k \paren {1 + x}\)

\(\ds \)

\(\ge\)

\(\ds \paren {1 + k x} \paren {1 + x}\)

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.

$\blacksquare$

Proof 2

Let $y = 1 + x$.

Then $y \ge 0$, and:

$\paren {1 + x}^n = 1 + \paren {y^n - 1}$

If $y \ge 1$, then by Sum of Geometric Sequence:

$\ds y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$

If $y < 1$, then by Sum of Geometric Sequence:

$\ds y^n - 1 = -\paren {1 - y} \sum_{k \mathop = 0}^{n - 1} y^k \ge -n \paren {1 - y} = n x$

Hence the result.

$\blacksquare$

Source of Name

This entry was named for Jacob Bernoulli.