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.
$\blacksquare$