Binomial theorem and the inequality $(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$ for $0<\alpha\le 1$

Question

Assume $0<\alpha\leq 1$ and $x>0$. Does the following inequality hold? $$(1-e^{-x})^{\alpha}\leq (1-\alpha e^{-x})$$ I know that the reverse inequality holds if $\alpha\ge 1$.

Answer 1

Using Bernoulli's Inequality

The reverse inequality, for $a\ge1$, is called Bernoulli's Inequality. It can be proven, in the case of integer exponents, using induction. This can be extended to the case of rational exponents, again using induction.

The inequality you mention above, for $0\lt a\le1$, follows immediately.

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Using the AM-GM Inequality

For $x\gt0$ and $0\le a\le1$, the AM-GM Inequality says that $$ 1^{1-a}x^a\le(1-a)1+ax $$ Substituting $x\mapsto1-x$ yields that for $x\lt1$ and $0\le a\le1$ $$ (1-x)^a\le1-ax $$

Answer 2

For $x > 0$, $0 < e^{-x} < 1$ (and this is a bijection between $(0,\infty)$ and $(0,1)$), so you can drop the exponentials and ask the same question about $$(1-y)^\alpha \leq 1-\alpha y$$ for $\alpha \in (0,1]$ and $y\in(0,1)$. Then, this should just be an exercise of convexity: consider the function $y\in(0,1) \mapsto (1-y)^\alpha$, for $\alpha \in (0,1]$.