# An exponentioal inequallity

• Aug 28th 2008, 08:08 PM
Andres Perez
An exponentioal inequallity
Show that for every $a\geq1$ and $x\geq 0$, $e^{ax}\geq 1+x^a$

thanks
• Aug 28th 2008, 08:32 PM
ThePerfectHacker
Quote:

Originally Posted by Andres Perez
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 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 such that $f(y) > f(x)$. But this is a contradiction because $f$ is increasing on $(o,\infty)$. Thus, $f\geq 0$.
• Aug 29th 2008, 01:48 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
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 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 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.
• Aug 29th 2008, 05:48 PM
JaneBennet
Quote:

Originally Posted by Andres Perez
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 (Evilgrin)
• Aug 29th 2008, 07:16 PM
NonCommAlg
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 $x^a.$
• Sep 8th 2008, 10:04 PM
bkarpuz
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 $a\geq 0.$