Editing Bernoulli's inequality (section)

=== Generalization of base ===
Instead of <math>(1+x)^n</math> the inequality holds also in the form <math>(1+x_1)(1+x_2)\dots(1+x_r) \geq 1+x_1+x_2 + \dots + x_r</math> where <math>x_1, x_2, \dots , x_r</math> are real numbers, all greater than <math>-1</math>, all with the same sign. Bernoulli's inequality is a special case when <math>x_1 = x_2 = \dots = x_r = x</math>. This generalized inequality can be proved by mathematical induction.

{{collapse top| title=Proof}}
In the first step we take <math>n=1</math>. In this case the inequality <math>1+x_1 \geq 1 + x_1</math> is obviously true.

In the second step we assume validity of the inequality for <math>r</math> numbers and deduce validity for <math>r+1</math> numbers.

We assume that<math display="block">(1+x_1)(1+x_2)\dots(1+x_r) \geq 1+x_1+x_2 + \dots + x_r</math>is valid. After multiplying both sides with a positive number <math>(x_{r+1} + 1)</math> we get:

<math>\begin{alignat}{2}
(1+x_1)(1+x_2)\dots(1+x_r)(1+x_{r+1}) \geq & (1+x_1+x_2 + \dots + x_r)(1+x_{r+1})
\\
\geq & (1+x_1+x_2+ \dots + x_r) \cdot 1 + (1+x_1+x_2+ \dots + x_r) \cdot x_{r+1}
\\
\geq & (1+x_1+x_2+ \dots + x_r) + x_{r+1} + x_1 x_{r+1} + x_2 x_{r+1} + \dots + x_r x_{r+1}
\\ \end{alignat}</math>

As <math>x_1, x_2, \dots x_r, x_{r+1}</math> all have the same sign, the products <math>x_1 x_{r+1}, x_2 x_{r+1}, \dots x_r x_{r+1}</math> are all positive numbers. So the quantity on the right-hand side can be bounded as follows:<math display="block">(1+x_1+x_2+ \dots + x_r) + x_{r+1} + x_1 x_{r+1} + x_2 x_{r+1} + \dots + x_r x_{r+1} \geq 1+x_1+x_2+ \dots + x_r + x_{r+1},</math>what was to be shown.
{{cob}}