Show that for every $\displaystyle a\geq1$ and $\displaystyle x\geq 0$, $\displaystyle e^{ax}\geq 1+x^a$

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- Aug 28th 2008, 07:08 PM #1

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- Aug 28th 2008, 07:32 PM #2

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Define $\displaystyle f(x) = e^{ax} - x^a - 1$ and show $\displaystyle f'>0$. Now $\displaystyle f(0)=0$. If there was $\displaystyle x>0$ such that $\displaystyle f(x)=0$ then since $\displaystyle f$ satisfies Rolle's theorem on $\displaystyle [0,x]$ it would mean there is a point $\displaystyle 0<y<x$ such that $\displaystyle f'(y) = 0$ and this is a contradiction. Thus, $\displaystyle f(x)\not = 0$ for $\displaystyle x\in (0,\infty)$. If there was $\displaystyle x<0$ such that $\displaystyle f(x)<0$ then since $\displaystyle f$ satisfies IVT it means there would be $\displaystyle 0<y<x$ such that $\displaystyle f(y) > f(x)$. But this is a contradiction because $\displaystyle f$ is increasing on $\displaystyle (o,\infty)$. Thus, $\displaystyle f\geq 0$.

- Aug 29th 2008, 12:48 PM #3

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well, if we prove that $\displaystyle f'(x) \geq 0$ for $\displaystyle x \geq 0$, then we're done because that means $\displaystyle f$ is increasing and thus: $\displaystyle f(x) \geq f(0)=0, \ \forall x \geq 0.$

the inequality is clearly true for x = 0. so we assume that x > 0. so the claim is: $\displaystyle \forall x > 0, \ \forall a \geq 1: \ \ f'(x)=a(e^{ax} - x^{a-1}) > 0. \ \ \ (1)$

suppose for now that a > 1. since $\displaystyle e^x > x,$ we have $\displaystyle x>\ln x.$ thus: $\displaystyle \frac{\ln x}{x} < 1 < \frac{a}{a-1}.$ hence $\displaystyle ax > (a-1)\ln x,$ which obviously holds

for a = 1 as well. therefore: $\displaystyle e^{ax} > e^{(a-1)\ln x}=x^{a-1},$ which proves (1). Q.E.D.

- Aug 29th 2008, 04:48 PM #4
$\displaystyle \forall\,x>0,a\ge1$, we have $\displaystyle x\ge\ln{x}$ and $\displaystyle a\ge a-1\ge0$

$\displaystyle \Rightarrow\ ax\ \ge\ (a-1)\ln{x}=\ln\left(x^{a-1}\right)$

$\displaystyle \Rightarrow\ e^{ax}\ \ge\ x^{a-1}$

$\displaystyle \Rightarrow\ ae^{ax}\ \ge\ ax^{a-1}$

This also holds for $\displaystyle x=0$; hence $\displaystyle ae^{ax}\ge ax^{a-1}$ for all $\displaystyle x\ge0$.

$\displaystyle \therefore\ \int_0^x{ae^{at}}\,dt\ \ge\ \int_0^x{at^{a-1}}\,dt$

$\displaystyle \Rightarrow\ e^{ax}-1\ \ge\ x^a$

QED

- Aug 29th 2008, 06:16 PM #5

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- Sep 8th 2008, 09:04 PM #6
A simple proof for the case $\displaystyle a\in\mathbb{N}$.

By the Taylor's expansion, we have $\displaystyle \exp\{x\}\geq 1+x$ since $\displaystyle x\geq0$ and by the binomial expansion, we have $\displaystyle (1+x)^{a}\geq 1+x^{a}.$

Using the fact that $\displaystyle \exp\{ax\}=\big(\exp\{x\}\big)^{a}$ and the inequalities above, we get the resired result since $\displaystyle t^{a}$ is increasing in $\displaystyle t$ when $\displaystyle a\geq 0.$