
#1
Nov908, 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. 



#2
Nov908, 12:23 PM

HW Helper
P: 5,004

This inequality is only true for a select range of xvalues, so what range are you supposed to prove this for?




#3
Nov908, 12:31 PM

P: 126

Sorry forgot to say: 0 < x < +∞




#4
Nov908, 12:36 PM

Mentor
P: 20,984

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. 



#5
Nov908, 12:53 PM

HW Helper
P: 5,004

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]?




#6
Nov908, 01:13 PM

P: 126

With: Right? But still, I can't think of anything but making that: And then replacing, but still nothing. 



#7
Nov908, 01:21 PM

HW Helper
P: 5,004

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)}{ba}=f'(c) \Rightarrow f(b)f(a)=(ba)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? 



#8
Nov908, 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? 



#10
Nov908, 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] 



#11
Nov908, 03:27 PM

HW Helper
P: 5,004

good, and what are the minimum and maximum values of [tex]\frac{\sin l}{3+\cos l}[/tex] for any real [itex]l[/itex]?




#12
Nov908, 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\geq1[/tex] Therefore: [tex]\frac{1}{2}\geq\frac{\sin l}{3+\cos l}\geq\frac{1}{4}[/tex] 



#13
Nov908, 03:53 PM

HW Helper
P: 5,004

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 xvalues, 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]? 



#14
Nov908, 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?




#16
Nov908, 04:22 PM

P: 126

OK, thanks a lot!



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