proof of Bernoulli’s inequality
Let $I$ be the interval $(1,\mathrm{\infty})$ and $f:I\to \mathbb{R}$ the function defined as:
$$f(x)={(1+x)}^{\alpha}1\alpha x$$ 
with $\alpha \in \mathbb{R}\setminus \{0,1\}$ fixed. Then $f$ is differentiable^{} and its derivative^{} is
$${f}^{\prime}(x)=\alpha {(1+x)}^{\alpha 1}\alpha ,\text{for all}x\in I,$$ 
from which it follows that ${f}^{\prime}(x)=0\iff x=0$.

1.
If $$ then $$ for all $x\in (0,\mathrm{\infty})$ and ${f}^{\prime}(x)>0$ for all $x\in (1,0)$ which means that $0$ is a global maximum^{} point for $f$. Therefore $$ for all $x\in I\setminus \{0\}$ which means that $$ for all $x\in (1,0)$.

2.
If $\alpha \notin [0,1]$ then ${f}^{\prime}(x)>0$ for all $x\in (0,\mathrm{\infty})$ and $$ for all $x\in (1,0)$ meaning that $0$ is a global minimum point for $f$. This implies that $f(x)>f(0)$ for all $x\in I\setminus \{0\}$ which means that ${(1+x)}^{\alpha}>1+\alpha x$ for all $x\in (1,0)$.
Checking that the equality is satisfied for $x=0$ or for $\alpha \in \{0,1\}$ ends the proof.
Title  proof of Bernoulli’s inequality^{} 

Canonical name  ProofOfBernoullisInequality 
Date of creation  20130322 12:38:14 
Last modified on  20130322 12:38:14 
Owner  danielm (240) 
Last modified by  danielm (240) 
Numerical id  6 
Author  danielm (240) 
Entry type  Proof 
Classification  msc 26D99 