proof of Bernoulli’s inequality

Let $I$ be the interval $(-1,\mathrm{\infty })$ and$f:I\to ℝ$ the function defined as:

$$f(x)={(1+x)}^{\alpha }-1-\alpha x$$

with $\alpha \in ℝ\setminus \{0,1\}$ 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 aglobal 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 globalminimum 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.