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Math Help - Exponential inequalities

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation Exponential inequalities

    Dear friends,

    I could not obtain proof for either one of the following inequalities:
    \mathrm{e}^{rx}\geq x+\frac{\mathrm{ln}(r)+1}{r}\text{ for }x\geq0\text{ and }r>0
    or
    \mathrm{e}^{rx}\geq x+\frac{\mathrm{ln}(\mathrm{e}r)}{r}\text{ for }x\geq0\text{ and }r>0.

    I tried to prove it by using their tangents, but it seems not to be so simple.
    Any ideas?

    Thanks.
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  2. #2
    MHF Contributor

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    There's probably a nicer proof, but here is one anyway:

    Let r>0. The derivative of the function f:x\mapsto e^{rx}-x is f':x\mapsto re^{rx}-1, which is equal to 0 at x=-\frac{\ln r}{r}, and respectively negative and positive at the left and right of this point. As a consequence, for every x\in\mathbb{R}, f(x)\geq f(-\frac{\ln r}{r})=\frac{1}{r}+\frac{\ln r}{r}, which is exactly what we want.

    Laurent.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Thanks, Laurent.
    You have any ideas for the second one?
    I guess that it is a little bit harder because your f function will be same for this inequality too.
    To be honest, I attempted to solve the latter one at first...
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  4. #4
    Super Member wingless's Avatar
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    What do you mean? That second one is the same as the first one.. Or are you talking about another question somewhere else?
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  5. #5
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by wingless View Post
    What do you mean? That second one is the same as the first one.. Or are you talking about another question somewhere else?
    Selam wingless, I am looking for answers for the following question, I think the proof for this one is different than the one given by Laurent:

    \mathrm{e}^{rx}\geq x+\frac{\mathrm{ln}(\mathrm{e}r)}{r}\text{ for }x\geq0\text{ and }r>0.
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  6. #6
    Super Member wingless's Avatar
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    Selam =)

    e^{rx} \geq x + \frac{\ln (er)}{r}

    e^{rx} \geq x + \frac{\ln e + \ln (r)}{r}

    e^{rx} \geq x + \frac{1 + \ln (r)}{r}

    So both inequalities are actually identical.

    Also notice that the question says 'or' between the inequalities...
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  7. #7
    Senior Member bkarpuz's Avatar
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    Thumbs up

    I have to say, that seems to be a joke :S
    How can i miss that thing!!!

    By the way, I put the 'or' there...
    Last edited by bkarpuz; September 8th 2008 at 11:47 AM.
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  8. #8
    Senior Member bkarpuz's Avatar
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    Angry

    Now, I see that I always had mistakes in the calculation.
    I again checked my way for the last time and saw that it works.

    Let me explain it below.

    Set f(x):=\mathrm{e}^{rx}, we clearly know that f(x)\geq f^{\prime}(x_{0})(x-x_{0})+f(x_{0})=f^{\prime}(x_{0})x+f(x_{0})-f^{\prime}(x_{0})x_{0} for every x_{0}\geq0 since f is convex.
    What I need to find is that
    f^{\prime}(x_{0})=1\qquad(1)
    and
    f(x_{0})-f^{\prime}(x_{0})x_{0}=f(x_{0})-x_{0}=\frac{1}{r}\mathrm{ln}(\mathrm{e}r).\qquad(2  )
    Solving (1), we get
    x_{0}=\frac{1}{r}\mathrm{ln}\bigg(\frac{1}{r}\bigg  )\qquad(3).
    Clearly, (3) satisfies (2), and thus the solution is complete.

    I really still don't know where I had mistakes all the time.
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