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Math Help - An exponentioal inequallity

  1. #1
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    An exponentioal inequallity

    Show that for every a\geq1 and x\geq 0, 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 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.
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  3. #3
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    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.
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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 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
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  5. #5
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    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.
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  6. #6
    Senior Member bkarpuz's Avatar
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    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.
    Last edited by bkarpuz; September 8th 2008 at 10:45 PM.
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