# Thread: Prove inequality using Mean Value Theorem

1. ## Prove inequality using Mean Value Theorem

Hi all!

I am desperately stuck. I need to prove this inequality: MathBin.net - Mean value theorem inequality.
On this page you also see my feeble attempts to do it. Have to hand it in tomorrow (same ol' story, I know ) and I don't know what to do!

Thanks for any help,

Ruben

2. pi/4 - x > 0 for every x in (-pi/2,pi/4), so...

However, I am afraid I don't understand what you are saying...
pi/4-x >= 0 for every x in (-pi/2, pi/4] right? But what use is that? I don't really understand how to solve these inequalities...

4. If $\displaystyle \frac{-\pi}{2}<x\le \frac{\pi}{4}$ then $\displaystyle \sec^2(x)\ge 1$.

Now apply the mean value theorem to $\displaystyle \tan(x)$.

5. You are here:

$\displaystyle \frac{\tan(\frac{\pi}{4})-\tan(x)}{\frac{\pi}{4}-x}\geq1$

Now, $\displaystyle \frac{\pi}{4} - x > 0$for every $\displaystyle x$ in $\displaystyle (-\frac{\pi}{2},\frac{\pi}{4})$, so you multiply by 1 without changing the sign.

Hence:

$\displaystyle 1-tan(x) \geq\frac{\pi}{4}-x}$

Can you proceed from here?

6. @Plato: thanks for your reply. I'm not used to the sec e.o. functions, so I prefer to use sin and cos. I already worked out the mean value theorem for tan(x) using sin and cos.
@Also sprach Zarathustra (nice name BTW ):
Thanks for your answer. I think I might have it now, but I know next to nothing about inequalities.
$\displaystyle \frac{1-\tan(x)}{\frac{\pi}{4}-x}& \geq 1$
$\displaystyle 1-\tan(x) & \geq \frac{\pi}{4}-x$
$\displaystyle x - \frac{\pi}{4} + 1 & \geq \tan(x)$
And of course $\displaystyle 1=\tan(\frac{\pi}{4})$, so:
$\displaystyle x - \frac{\pi}{4} + 1 &= x - \frac{\pi}{4} + \tan(\frac{\pi}{4}) \geq \tan(x)$
I only don't know whether this is correct. Also, why can you only multiply by 1 without changing the sign if $\displaystyle \frac{\pi}{4} - x > 0$?
You have been a great help so far, thank you!

7. Originally Posted by Ruben
@Plato: thanks for your reply. I'm not used to the sec e.o. functions, so I prefer to use sin and cos. I already worked out the mean value theorem for tan(x) using sin and cos.!
Pray tell us why you put the mean value theorem in the title of this posting?
After all the mean value theorem is about derivatives.
The derivative of $\displaystyle \tan(x)$ is the $\displaystyle \sec^2(x)$.

8. Well, I had to use the mean value theorem to prove it. (That was given in the assignment)
Also, note that:
$\displaystyle \tan(x) = \frac{\sin(x)}{\cos(x)}$
$\displaystyle [\frac{\sin(x)}{\cos(x)}]^{\prime} =\frac{2}{\cos(2x)+1}$
Using the quotient rule and some simplification...

9. Originally Posted by Ruben
Well, I had to use the mean value theorem to prove it. (That was given in the assignment)
Also, note that:
$\displaystyle \tan(x) = \frac{\sin(x)}{\cos(x)}$
$\displaystyle [\frac{\sin(x)}{\cos(x)}]^{\prime} =\frac{2}{\cos(2x)+1}$
Using the quotient rule and some simplification...
I think I can say that is a unique way of doing that without usage error.
I have taught calculus for almost forty-five years, but never have I seen that.