The Weierstrass Inequality is a useful result in mathematics that helps us estimate products of numbers that are slightly bigger than 1.
Statement of the Inequality
If
\[ x_1, x_2, x_3, \dots, x_n \geq -1 \]Then
\[ (1 + x_1)(1 + x_2)(1 + x_3)\dots(1 + x_n) \geq 1 + x_1 + x_2 + x_3 + \dots + x_n \]
In simple words:
When you multiply numbers of the form \( (1 + x_i) \), the result is always greater than or equal to \( 1 + \) the sum of those \( x_i \)’s.
When you multiply numbers of the form \( (1 + x_i) \), the result is always greater than or equal to \( 1 + \) the sum of those \( x_i \)’s.
Simple Example
Let
\[ x_1 = 0.2, \quad x_2 = 0.3 \]Left side:
\[ (1 + 0.2)(1 + 0.3) = (1.2)(1.3) = 1.56 \]Right side:
\[ 1 + 0.2 + 0.3 = 1.5 \]Since \( 1.56 \geq 1.5 \), the inequality is true.
Why Is It True?
For two numbers:
\[ (1 + x_1)(1 + x_2) \]Expanding:
\[ = 1 + x_1 + x_2 + x_1 x_2 \]Since \( x_1, x_2 \geq -1 \), the extra term \( x_1 x_2 \) makes the expression larger (or at least not smaller). Therefore,
\[ 1 + x_1 + x_2 + x_1 x_2 \geq 1 + x_1 + x_2 \]When Does Equality Occur?
Equality happens when one or more of the \( x_i \) values are zero, so that all extra product terms vanish.
Where Is It Used?
- Algebraic estimations
- Bounding expressions
- Series and product approximations
- Advanced calculus and analysis
Final Takeaway
\[ (1 + x_1)(1 + x_2)\dots(1 + x_n) \geq 1 + \sum_{i=1}^{n} x_i \]The product of terms of the form \( (1 + x_i) \) is always at least \( 1 \) plus their sum — provided each \( x_i \geq -1 \).