Bernoulli's Inequality

analysis

Definition

Bernoulli's Inequality

For every real number $x\ge -1$ and every natural number $n\in \mathbb{N}$, the following inequality holds:

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

If $n\ge 2$ and $x>-1,x\ne 0$, the inequality is strict:$(1+x{)}^n>1+nx$.

Proof by Induction

Bernoulli's Inequality

We use induction on $n$.

Base case ($n=0$): >$(1+x{)}^0=1\ge 1+0x=1$. The base case holds.

Inductive step:
Assume $(1+x{)}^k\ge 1+kx$ for some $k\in \mathbb{N}$. We want to show it holds for $k+1$.

$$\begin{array}{rl}(1+x{)}^{k+1}& =(1+x{)}^k(1+x)\\ & \ge (1+kx)(1+x)\text{(by inductive hypothesis, since }1+x\ge 0)\\ & =1+x+kx+k{x}^2\\ & =1+(k+1)x+k{x}^2\\ & \ge 1+(k+1)x\text{(since }k{x}^2\ge 0)\end{array}$$

Thus, the statement holds for all $n\in \mathbb{N}$ by induction.