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Thread: How to prove this Inequality?

  1. #1
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    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!
    Last edited by mr fantastic; Oct 24th 2010 at 11:34 AM. Reason: Title
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rectangle View Post
    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

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

    Now, find the contradiction
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  3. #3
    MHF Contributor Unknown008's Avatar
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    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...
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Unknown008 View Post

    EDIT: A little too late...
    You have a different approach, so it's fine.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    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)..
    :-/
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rectangle View Post
    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.
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  7. #7
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    Thank you so much!!!
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    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" ?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rectangle View Post
    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?
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    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...
    :-/
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  11. #11
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    Quote Originally Posted by rectangle View Post
    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
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  12. #12
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    Thank you so much.
    I am such an idiot.
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  13. #13
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    No, you are not.
    You persisted. That's how to learn. Keep it up.
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  14. #14
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by rectangle View Post
    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.
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