# proof of Bernoulli’s inequality

Let $I$ be the interval $(-1,\infty)$ and $f:I\rightarrow\mathbb{R}$ 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

 $f^{\prime}(x)=\alpha(1+x)^{\alpha-1}-\alpha,\mbox{ for all }x\in I,$

from which it follows that $f^{\prime}(x)=0\Leftrightarrow x=0$.

1. 1.

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

2. 2.

If $\alpha\notin[0,1]$ then $f^{\prime}(x)>0$ for all $x\in(0,\infty)$ and $f^{\prime}(x)<0$ 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 ProofOfBernoullisInequality 2013-03-22 12:38:14 2013-03-22 12:38:14 danielm (240) danielm (240) 6 danielm (240) Proof msc 26D99