# Thread: How to prove this Inequality?

1. ## How to prove this Inequality?

Hi...

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

Thank you!

2. Originally Posted by rectangle
Hi...

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

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

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

3. 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...

4. Originally Posted by Unknown008

EDIT: A little too late...
You have a different approach, so it's fine.

5. 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

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

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)..
:-/

6. 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.

7. Thank you so much!!!

8. 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" ?

9. 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?

10. 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...
:-/

11. 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

12. Thank you so much.
I am such an idiot.

13. No, you are not.
You persisted. That's how to learn. Keep it up.

14. 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.