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Help proving an inequality

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to prove an inequality for 0 < x < +∞.

    2. Relevant equations
    [​IMG]

    3. The attempt at a solution
    I guess it must be something with the mean value theorem, but I can't find what it is.

    Thanks for your help.

    Edit: Added range for x.
     
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2

    gabbagabbahey

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    This inequality is only true for a select range of x-values, so what range are you supposed to prove this for?
     
  4. Nov 9, 2008 #3
    Sorry forgot to say: 0 < x < +∞
     
  5. Nov 9, 2008 #4

    Mark44

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    I don't see where restricting x to the positive reals does anything for you.

    2 <= 3 + cos(x) <= 4, for all real x, so ln(3 + cos(x)) is defined for all real x, as well.
    Taking logs, the inequality above becomes
    ln(2) <= ln(3 + cos(x)) <= ln(4) = 2ln(2)

    I don't see how the lower and upper bounds of the original inequality figure in, though.
     
  6. Nov 9, 2008 #5

    gabbagabbahey

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    Hint: Let [itex]f(x)=\ln (3+ \cos x)[/itex]...what does the mean value theorem have to say about [itex]f(x)-f(0)[/itex]?
     
  7. Nov 9, 2008 #6
    [​IMG]

    With: [​IMG]

    Right?


    But still, I can't think of anything but making that:
    [​IMG]

    And then replacing, but still nothing.
     
    Last edited: Nov 9, 2008
  8. Nov 9, 2008 #7

    gabbagabbahey

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    Perhaps we should go over the mean value theorem first...it says that for some [itex]c \in [a,b] [/itex]
    , the following will hold true:

    [tex]\frac{f(b)-f(a)}{b-a}=f'(c) \Rightarrow f(b)-f(a)=(b-a)f'(c)[/tex]

    So looking at [itex]f(x)-f(0)[/itex] means [itex]a=0[/itex] and [itex]b=x[/itex] in the above relation right?....what does that give you?
     
  9. Nov 9, 2008 #8
    I thought [tex]b = 3+cos x[/tex] and [tex]a = 3+cos 0 = 4[/tex] :rolleyes:

    So for: [tex]0 \leq l \leq x[/tex] we have:
    [tex]ln(3+cos x) - ln(4) = \frac{x}{l}[/tex]
    Is that right so far?
     
  10. Nov 9, 2008 #9

    gabbagabbahey

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    What happened to [itex]f'(l)[/itex]?
     
  11. Nov 9, 2008 #10
    OK, I forgot that it's not [tex]ln(l)[/tex] but rather [tex]ln(3+\cos l)[/tex]
    So the derivative would be:
    [tex]-\frac{\sin l}{3+\cos l}[/tex]
    So finally:
    [tex]\ln(3+\cos x)-\ln 4 = \frac{-x\sin l}{3+\cos l}[/tex]
     
  12. Nov 9, 2008 #11

    gabbagabbahey

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    good, and what are the minimum and maximum values of [tex]\frac{-\sin l}{3+\cos l}[/tex] for any real [itex]l[/itex]?
     
  13. Nov 9, 2008 #12
    The only way I can think of for doing that is:
    [tex]\frac{1}{2}\geq\frac{1}{3+\cos l}\geq\frac{1}{4}[/tex]

    And:
    [tex]1\geq-\sin l\geq-1[/tex]

    Therefore:
    [tex]\frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{4}[/tex]
     
  14. Nov 9, 2008 #13

    gabbagabbahey

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    Close, the actual relationship is:

    [tex]
    \frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{2}
    [/tex]

    (since -1/4 is actually greater than -1/2 not less than)

    Given this, and the fact that you are restricted to positive x-values, what are the restrictions on [tex]\frac{-x\sin l}{3+\cos l}[/tex] ?

    And hence what are the maximum and minimum values of [itex]f(x)-f(0)[/itex]?
     
    Last edited: Nov 9, 2008
  15. Nov 9, 2008 #14
    OK, I think I got it now, but am I supposed to know that [tex]1\leq\ln4\leq\frac{3}{2}[/tex] in order to answer or is there any way around it?
     
  16. Nov 9, 2008 #15

    gabbagabbahey

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    I think that's all there is to it.
     
  17. Nov 9, 2008 #16
    OK, thanks a lot!
     
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