# Help proving an inequality

by springo
Tags: inequality, proving
 P: 126 1. The problem statement, all variables and given/known data I need to prove an inequality for 0 < x < +∞. 2. Relevant equations $1-\frac{x}{2} \leq Ln(3+cos x) \leq \frac{3}{2}+\frac{x}{2}$ 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.
 HW Helper P: 5,004 This inequality is only true for a select range of x-values, so what range are you supposed to prove this for?
 P: 126 Sorry forgot to say: 0 < x < +∞
Mentor
P: 19,671

## 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.
 HW Helper P: 5,004 Hint: Let $f(x)=\ln (3+ \cos x)$...what does the mean value theorem have to say about $f(x)-f(0)$?
 P: 126 $ln(3+cos x) - ln(4) = \frac{cos x-1}{l}$ With: $3+cos x \leq l \leq 4$ Right? But still, I can't think of anything but making that: $2 \leq l \leq 4$ And then replacing, but still nothing.
 HW Helper P: 5,004 Perhaps we should go over the mean value theorem first...it says that for some $c \in [a,b]$ , the following will hold true: $$\frac{f(b)-f(a)}{b-a}=f'(c) \Rightarrow f(b)-f(a)=(b-a)f'(c)$$ So looking at $f(x)-f(0)$ means $a=0$ and $b=x$ in the above relation right?....what does that give you?
 P: 126 I thought $$b = 3+cos x$$ and $$a = 3+cos 0 = 4$$ So for: $$0 \leq l \leq x$$ we have: $$ln(3+cos x) - ln(4) = \frac{x}{l}$$ Is that right so far?
 HW Helper P: 5,004 What happened to $f'(l)$?
 P: 126 OK, I forgot that it's not $$ln(l)$$ but rather $$ln(3+\cos l)$$ So the derivative would be: $$-\frac{\sin l}{3+\cos l}$$ So finally: $$\ln(3+\cos x)-\ln 4 = \frac{-x\sin l}{3+\cos l}$$
 HW Helper P: 5,004 good, and what are the minimum and maximum values of $$\frac{-\sin l}{3+\cos l}$$ for any real $l$?
 P: 126 The only way I can think of for doing that is: $$\frac{1}{2}\geq\frac{1}{3+\cos l}\geq\frac{1}{4}$$ And: $$1\geq-\sin l\geq-1$$ Therefore: $$\frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{4}$$
 HW Helper P: 5,004 Close, the actual relationship is: $$\frac{1}{2}\geq\frac{-\sin l}{3+\cos l}\geq-\frac{1}{2}$$ (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 $$\frac{-x\sin l}{3+\cos l}$$ ? And hence what are the maximum and minimum values of $f(x)-f(0)$?
 P: 126 OK, I think I got it now, but am I supposed to know that $$1\leq\ln4\leq\frac{3}{2}$$ in order to answer or is there any way around it?
HW Helper
P: 5,004
 Quote by springo OK, I think I got it now, but am I supposed to know that $$1\leq\ln4\leq\frac{3}{2}$$ in order to answer or is there any way around it?
I think that's all there is to it.
 P: 126 OK, thanks a lot!

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