# How to prove this Inequality?

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• Oct 24th 2010, 10:06 AM
rectangle
How to prove this Inequality?
Hi...

I didn't understand this at all...
could someone solve it and explain me how he did it?

http://img827.imageshack.us/img827/1821/65184366.jpg

Thank you!
• Oct 24th 2010, 10:33 AM
Jhevon
Quote:

Originally Posted by rectangle
Hi...

I didn't understand this at all...
could someone solve it and explain me how he did it?

http://img827.imageshack.us/img827/1821/65184366.jpg

Thank you!

Given the ranges of $\displaystyle \displaystyle \alpha \text{ and } \beta$, clearly both $\displaystyle \displaystyle \tan \alpha , ~ \tan \beta > 0$, and so, their sum will be positive also. That is, $\displaystyle \displaystyle 0 < \tan \alpha + \tan \beta$.

Now it remains only to show that $\displaystyle \displaystyle \tan \alpha + \tan \beta < 1$. Assume, to the contrary, that $\displaystyle \displaystyle \tan \alpha + \tan \beta \ge 1$.

Since $\displaystyle \displaystyle \alpha = \frac {\pi}4 - \beta$, this is the same as saying

$\displaystyle \displaystyle \tan \left( \frac {\pi}4 - \beta \right) + \tan \beta \ge 1$

Now, find the contradiction
• Oct 24th 2010, 10:46 AM
Unknown008
Introduce tan on both sides of the given equation:

$\displaystyle \tan(\alpha + \beta) = \tan \dfrac{\pi}{4}$.

This becomes:

$\displaystyle \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}= 1$

$\displaystyle \tan\alpha + \tan\beta = 1 - \tan\alpha\tan\beta$

Now, since $\displaystyle \tan\alpha$ and $\displaystyle \tan\beta$ are positive, $\displaystyle \tan\alpha\tan\beta$ is also positive.

So, $\displaystyle 1 - \tan\alpha\tan\beta$ is positive. and less than 1

And the minimum of $\displaystyle \tan\alpha + \tan\beta$ is zero, which occurs only when $\displaystyle \tan\alpha = 0$ or $\displaystyle \tan\beta = 0$, which is not the case.

Hence, we get

$\displaystyle 0 < \tan\alpha + \tan\beta < 1$

EDIT: A little too late...
• Oct 24th 2010, 10:51 AM
Jhevon
Quote:

Originally Posted by Unknown008

EDIT: A little too late...

You have a different approach, so it's fine.
• Oct 24th 2010, 10:54 AM
rectangle
Quote:

Originally Posted by Jhevon
Given the ranges of $\displaystyle \displaystyle \alpha \text{ and } \beta$, clearly both $\displaystyle \displaystyle \tan \alpha , ~ \tan \beta > 0$, and so, their sum will be positive also. That is, $\displaystyle \displaystyle 0 < \tan \alpha + \tan \beta$.

Now it remains only to show that $\displaystyle \displaystyle \tan \alpha + \tan \beta < 1$. Assume, to the contrary, that $\displaystyle \displaystyle \tan \alpha + \tan \beta \ge 1$.

Since $\displaystyle \displaystyle \alpha = \frac {\pi}4 - \beta$, this is the same as saying

$\displaystyle \displaystyle \tan \left( \frac {\pi}4 - \beta \right) + \tan \beta \ge 1$

Now, find the contradiction

Hi...
I understood the first part where both of the tan is bigger than 0,
but didn't understand how to prove the second part (smaller than 1)..
:-/
• Oct 24th 2010, 11:29 AM
Jhevon
Quote:

Originally Posted by rectangle
Hi...
I understood the first part where both of the tan is bigger than 0,
but didn't understand how to prove the second part (smaller than 1)..
:-/

see post #3. the solution there is more elementary, and you should be able to follow it.
• Oct 24th 2010, 11:37 AM
rectangle
Thank you so much!!!
• Oct 24th 2010, 12:16 PM
rectangle
Quote:

Originally Posted by Unknown008
Introduce tan on both sides of the given equation:

$\displaystyle \tan(\alpha + \beta) = \tan \dfrac{\pi}{4}$.

This becomes:

$\displaystyle \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}= 1$

$\displaystyle \tan\alpha + \tan\beta = 1 - \tan\alpha\tan\beta$

Now, since $\displaystyle \tan\alpha$ and $\displaystyle \tan\beta$ are positive, $\displaystyle \tan\alpha\tan\beta$ is also positive.

So, $\displaystyle 1 - \tan\alpha\tan\beta$ is positive. and less than 1

And the minimum of $\displaystyle \tan\alpha + \tan\beta$ is zero, which occurs only when $\displaystyle \tan\alpha = 0$ or $\displaystyle \tan\beta = 0$, which is not the case.

Hence, we get

$\displaystyle 0 < \tan\alpha + \tan\beta < 1$

EDIT: A little too late...

Thank you...can you explain me how can you prove that " $\displaystyle 1 - \tan\alpha\tan\beta$ is positive. and less than 1" ?
• Oct 24th 2010, 01:37 PM
Jhevon
Quote:

Originally Posted by rectangle
Thank you...can you explain me how can you prove that " $\displaystyle 1 - \tan\alpha\tan\beta$ is positive. and less than 1" ?

He did explain. What didn't you get?
• Oct 24th 2010, 01:48 PM
rectangle
Quote:

Originally Posted by Jhevon
He did explain. What didn't you get?

If tan alpha and tan beta = tan45 then it equals to 1.
I do understand that i was given a face that
alpha + beta = pi/4
so it will be always smaller than 1, but i am not sure how to write it...
:-/
• Oct 24th 2010, 03:12 PM
Archie Meade
Quote:

Originally Posted by rectangle
Thank you...can you explain me how can you prove that " $\displaystyle 1 - \tan\alpha\tan\beta$ is positive. and less than 1" ?

Although you are given that both $\displaystyle 0<\alpha<\frac{{\pi}}{2}$ and $\displaystyle 0<\beta<\frac{{\pi}}{2}$

since $\displaystyle \alpha+\beta=\frac{{\pi}}{4}$

then both angles are in fact $\displaystyle <\frac{{\pi}}{4}$

hence $\displaystyle Tan\alpha<1$ and $\displaystyle Tan\beta<1$

Two positive fractions that are <1 when multiplied give a positive result <1

Hence, subtracting that result from 1 gives a value <1
• Oct 24th 2010, 03:50 PM
rectangle
Thank you so much.
I am such an idiot.
• Oct 24th 2010, 03:57 PM
Archie Meade
No, you are not.
You persisted. That's how to learn. Keep it up.
• Oct 24th 2010, 08:37 PM
Unknown008
Quote:

Originally Posted by rectangle
If tan alpha and tan beta = tan45 then it equals to 1.
I do understand that i was given a face that
alpha + beta = pi/4
so it will be always smaller than 1, but i am not sure how to write it...
:-/

What is not true here, is that you said that:

$\displaystyle \tan \alpha + \tan \beta = \tan45$

What is true is

$\displaystyle \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}= 1$

Since $\displaystyle \tan \alpha + \tan \beta$ is divided by something, it is not equal to 1.