Bernoulli's Inequality/Corollary/General Result

Theorem

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

The proof proceeds by induction.

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

$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$

$\map P 0$ is the case:

\(\ds \prod_{j \mathop = 1}^0 \paren {1 - a_j}\)

\(=\)

\(\ds 1\)

Definition of Vacuous Product

\(\ds \)

\(\ge\)

\(\ds 1 - 0\)

\(\ds \)

\(=\)

\(\ds 1 - \sum_{j \mathop = 1}^0 a^j\)

Definition of Vacuous Summation

Thus $\map P 0$ is seen to hold.

Basis for the Induction

$\map P 1$ is the case:

\(\ds \prod_{j \mathop = 1}^1 \paren {1 - a_j}\)

\(=\)

\(\ds 1 - a_j\)

\(\ds \)

\(=\)

\(\ds 1 - \sum_{j \mathop = 1}^1 a^j\)

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\ds \prod_{j \mathop = 1}^k \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^k a^j$

from which it is to be shown that:

$\ds \prod_{j \mathop = 1}^{k + 1} \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^{k + 1} a^j$

Induction Step

This is the induction step:

\(\ds \prod_{j \mathop = 1}^{k + 1} \paren {1 - a_j}\)

\(=\)

\(\ds \paren {\prod_{j \mathop = 1}^k \paren {1 - a_j} } \paren {1 - a_{k + 1} }\)

Definition of Continued Product

\(\ds \)

\(\ge\)

\(\ds \paren {1 - \sum_{j \mathop = 1}^k a^j} \paren {1 - a_{k + 1} }\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds 1 - \sum_{j \mathop = 1}^k a^j - a_{k + 1} + a_{k + 1} \sum_{j \mathop = 1}^k a^j\)

\(\ds \)

\(=\)

\(\ds 1 - \sum_{j \mathop = 1}^{k + 1} a^j + a_{k + 1} \sum_{j \mathop = 1}^k a^j\)

\(\ds \)

\(\ge\)

\(\ds 1 - \sum_{j \mathop = 1}^{k + 1} a^j\)

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

Therefore:

$\ds \forall n \in \Z_{\ge 0}: \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $27$