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Help proving an inequality

by springo
Tags: inequality, proving
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springo
#1
Nov9-08, 11:28 AM
P: 126
1. The problem statement, all variables and given/known data
I need to prove an inequality for 0 < x < +∞.

2. Relevant equations


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.
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gabbagabbahey
#2
Nov9-08, 12:23 PM
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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?
springo
#3
Nov9-08, 12:31 PM
P: 126
Sorry forgot to say: 0 < x < +∞

Mark44
#4
Nov9-08, 12:36 PM
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P: 21,216
Help proving an inequality

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.
gabbagabbahey
#5
Nov9-08, 12:53 PM
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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]?
springo
#6
Nov9-08, 01:13 PM
P: 126


With:

Right?


But still, I can't think of anything but making that:


And then replacing, but still nothing.
gabbagabbahey
#7
Nov9-08, 01:21 PM
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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?
springo
#8
Nov9-08, 02:17 PM
P: 126
I thought [tex]b = 3+cos x[/tex] and [tex]a = 3+cos 0 = 4[/tex]

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?
gabbagabbahey
#9
Nov9-08, 02:50 PM
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What happened to [itex]f'(l)[/itex]?
springo
#10
Nov9-08, 03:22 PM
P: 126
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]
gabbagabbahey
#11
Nov9-08, 03:27 PM
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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]?
springo
#12
Nov9-08, 03:47 PM
P: 126
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]
gabbagabbahey
#13
Nov9-08, 03:53 PM
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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]?
springo
#14
Nov9-08, 04:12 PM
P: 126
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?
gabbagabbahey
#15
Nov9-08, 04:15 PM
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Quote Quote by springo View Post
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?
I think that's all there is to it.
springo
#16
Nov9-08, 04:22 PM
P: 126
OK, thanks a lot!


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