Bernoulli's inequality

Theorem.

If  $\alpha\geq 1$ then \[ (1+x)^{\alpha}\geq 1+\alpha x, \quad x\geq -1. \]

Define the function \[ f(x):=(1+x)^{\alpha}-1-\alpha x, \quad x> -1. \] Then \[ f'(x)=\alpha (1+x)^{\alpha-1}-\alpha=\alpha\left( (1+x)^{\alpha-1}-1 \right). \]

Properties of $f(x)$

	$(-1,0)$	$0$	$(0,\infty)$
$f’(x)$	$-$	$0$	$+$
$f(x)$	strictly decreasing	local (global) minimum $f(0)=0$	strictly increasing

From this table it follows that $f(x)\geq 0$ what we wanted to prove.  $\blacksquare$

Application.

A remarkable sequence Theorem 7.