Bernoulli's Inequality/Proof 2

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Theorem

Let $x \in \R$ be a real number such that $x > -1$.

Let $n \in \Z_{\ge 0}$ be a positive integer.


Then:

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


Proof

Let $y = 1 + x$.

Then $y \ge 0$, and:

$\left({1 + x}\right)^n = 1 + \left({y^n - 1}\right)$

If $y \ge 1$, then by Sum of Geometric Progression:

$\displaystyle y^n - 1 = \left({y - 1}\right) \sum_{k \mathop = 0}^{n-1} y^k \ge n \left({y - 1}\right) = nx$

If $y < 1$, then by Sum of Geometric Progression:

$\displaystyle y^n - 1 = -\left({1 - y}\right) \sum_{k \mathop = 0}^{n-1} y^k \ge -n \left({1 - y}\right) = nx$

Hence the result.

$\blacksquare$


Source of Name

This entry was named for Jacob Bernoulli.