Proof of Bernoulli's Inequality
Theorem
If we call it $\alpha > 0$, for all $x \in [ - 1, \infty )$, the following two inequalities hold:
- [1]: $\alpha \in (0, 1] \implies (1 + x )^{\alpha } \le 1 + \alpha x $
- [2] $\alpha \in (1, \infty] \implies (1 + x )^{\alpha } \ge 1 + \alpha x $
Explanation
Looking closely at the shape of the inequalities, although it depends on the size of $\alpha$, one side involves multiplication and the other involves exponentiation. Of course, depending on the conditions, being able to switch one uncomfortable calculation to a preferred side is very good.
Proof
Strategy: It’s almost over just by using the mean value theorem. The proof of [2] is **omitted since it is almost the same as [1].
[1]
If the function $f : [ -1 , \infty) \to \mathbb{R}$ is defined as $f(t) := t^{\alpha}$, then $f’ (t) = \alpha t^{ \alpha -1 }$. By the mean value theorem, there exists $c$ between $1$ and $1 + x$ that satisfies $$ \begin{align} f(1+x) - f(1) = \alpha x c^{\alpha - 1} \label{1} \end{align} $$
Case 1. $0 < x$
Since $c \in (1, 1+ x)$, it follows that $c>1$ and since $\alpha - 1 \le 0$ $$ c^{\alpha -1} \le 1 $$ Multiplying both sides by $x > 0$, we get $$ x c^{\alpha -1} \le x $$
Case 2. $-1 \le x \le 0$
Since $c \in (1 + x, 1)$, it follows that $c \le 1$ and since $\alpha - 1 \le 0$ $$ c^{\alpha - 1} \ge 1 $$ Multiplying both sides by $x \le 0$, we get $$ x c^{\alpha -1} \le x $$ In any case, according to $\eqref{1}$, $$ \begin{align*} \displaystyle (1 +x)^{\alpha} =& f(1+x) \\ =& f(1) + \alpha x c^{ \alpha - 1 } \\ \le & f(1) + \alpha x \\ =& 1 + \alpha x \end{align*} $$
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Meanwhile, for all $x > 0$, since $1 + x < e^{x}$, we obtain the following corollary for $e \in [ 1, \infty )$.
Corollary
$$(1 + x )^{ \alpha } \le e^{ x \alpha }$$