Bernoulli's Inequality/Corollary/Proof 2

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:

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

Proof

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.

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction: Exercise $9$