An exponentioal inequallity

Andres Perez · #1 $Old$ 08-28-2008, 09:08 PM

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

thanks

ThePerfectHacker · #2 $Old$ 08-28-2008, 09:32 PM

Quote:

Originally Posted by Andres Perez $View Post$

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

thanks

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

NonCommAlg · #3 $Old$ 08-29-2008, 02:48 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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

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

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

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

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

JaneBennet · #4 $Old$ 08-29-2008, 06:48 PM

Quote:

Originally Posted by Andres Perez $View Post$

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

thanks

$\forall\,x>0,a\ge1$ , we have $x\ge\ln{x}$ and $a\ge a-1\ge0$

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

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

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

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

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

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

QED

NonCommAlg · #5 $Old$ 08-29-2008, 08:16 PM

try this improved version of Andres Perez's inequality: prove that $\forall x >0, \ \forall a \geq 1: \ e^{ax} > 1 + x^{2a}.$

so this time on the RHS we have $x^{2a}$ instead of $x^a.$

bkarpuz · #6 $Old$ 09-08-2008, 11:04 PM

A simple proof for the case $a\in\mathbb{N}$ .
By the Taylor's expansion, we have $\exp\{x\}\geq 1+x$ since $x\geq0$ and by the binomial expansion, we have $(1+x)^{a}\geq 1+x^{a}.$
Using the fact that $\exp\{ax\}=\big(\exp\{x\}\big)^{a}$ and the inequalities above, we get the resired result since $t^{a}$ is increasing in $t$ when $a\geq 0.$