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Math Help - derivative inequality

  1. #1
    Junior Member miss_lolitta's Avatar
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    derivative inequality

    can someone help me??!

    I have this problem..

    if  x(t)^{'} is a nonnegative and decreasing function and

     x(t) is a positive and nondecreasing

    so..

     \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t} for all  t\geqslant b ??!

    I tried to prove this problem but at this step

     x(t)-x(b)\geqslant(t-b)x(t)^{'}

    I don't know how to get the rest of proof from this step..

    can someone help me to get the rest of proof??

    thanks for any help
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  2. #2
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    What is x(t), what is x'(t), on what interval are they defined? What is b?
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  3. #3
    Junior Member miss_lolitta's Avatar
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    moredetial

    if  x^{'}(t) is a nonnegative and decreasing function
    and  x(t) is a positive and nondecreasing on [b,t]

    then

    x(t)-x(b)=\int^{t}_{b}{x^{'}(s)ds}

    since  x^{'}(t) is decreasing

     x^{'}(t)\leqslant x^{'}(s)


    this implies

     x(t)-x(b)\geqslant x^{'}(t) \int^{t}_{b}{ds}

    so

    x(t)-x(b)\geqslant(t-b)x^{'}(t)

    i.e.

     \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t-b} .....(*)

    now,I just want to get

     \frac{x^{'}(t)}{x(t)}\leqslant \frac{1}{t} ??!

    I hope that's clear
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