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Thread: An exponentioal inequallity

  1. #1
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    An exponentioal inequallity

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

    thanks
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  2. #2
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    Quote Originally Posted by Andres Perez View Post
    Show that for every $\displaystyle a\geq1$ and $\displaystyle x\geq 0$, $\displaystyle e^{ax}\geq 1+x^a$

    thanks
    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, $\displaystyle f\geq 0$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    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, $\displaystyle f\geq 0$.
    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.
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Andres Perez View Post
    Show that for every $\displaystyle a\geq1$ and $\displaystyle x\geq 0$, $\displaystyle e^{ax}\geq 1+x^a$

    thanks
    $\displaystyle \forall\,x>0,a\ge1$, 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
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  5. #5
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    try this improved version of Andres Perez's inequality: prove that $\displaystyle \forall x >0, \ \forall a \geq 1: \ e^{ax} > 1 + x^{2a}.$

    so this time on the RHS we have $\displaystyle x^{2a}$ instead of $\displaystyle x^a.$
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  6. #6
    Senior Member bkarpuz's Avatar
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    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 $\displaystyle a\geq 0.$
    Last edited by bkarpuz; Sep 8th 2008 at 09:45 PM.
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