01 - Bernoulli's Inequality

Bernoulli's Inequality

Let $x$ be a real number with $x>-.1$
Let $n$ be a positive integer then

$$(1+x{)}^n\ge 1+nx$$

Proof

By induction on $n$
Fix $x>-1$

Base Case: $n=1$

$$(1+x{)}^1=1+x=1+1\cdot x=1+nx$$

So statement holds true for $n=1$

Inductive Hypothesis:
Suppose the statement holds true for $n=k$ (for some $k\ge 1$) hence

$$(1+x{)}^k\ge 1+kx$$

Note that $1+x>0$, and $n{x}^2>0$ (since $n>0$ and $x^2\ge 0$ by Results on Inequalities (5))
Inductive Step: $n=k+1$

$$\begin{array}{rlrl}(1+x{)}^{n+1}& =(1+x)(1+x{)}^n& & (\text{ by defintion})\\ & \ge (1+x)(1+nx)& & (\text{by inductive hypothesis})\\ & =1+(n+1)x+n{x}^2& & (\text{properties of arithmetic})\\ & \ge 1+(n+1)x& & (\text{since }n{x}^2>0)\end{array}$$