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 $\alpha\ge 1$.
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In fact, I believe the reverse inequality holds – Ben Grossmann Nov 25 '15 at 14:55
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2Reverse inequality hold if $\alpha >1$. – sudha pandey Nov 25 '15 at 15:02
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Ah, thanks sudha – Ben Grossmann Nov 25 '15 at 16:51
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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 $\alpha \in (0,1]$.
Clement C.
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No. Have you even read my answer? The word "convexity" should be a hint. If a function is concave, then it is _____ its asymptotes. And you should know how to prove a smooth and easily differentiable function is convex or concave. If you are looking for a full-fledged, complete and detailed answer without your actually having to do anything, then indeed this is not what I gave. – Clement C. Nov 25 '15 at 15:12
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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.
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 $0\le a\le1$ $$ (1-x)^a\le1-ax $$
