Results 1 to 14 of 14

Math Help - How to prove this Inequality?

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    11

    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; October 24th 2010 at 12:34 PM. Reason: Title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    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 \alpha \text{ and } \beta, clearly both \displaystyle \tan \alpha , ~ \tan \beta > 0, and so, their sum will be positive also. That is, \displaystyle 0 < \tan \alpha + \tan \beta.

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

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

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

    Now, find the contradiction
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Introduce tan on both sides of the given equation:

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

    This becomes:

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

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

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

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

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

    Hence, we get

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

    EDIT: A little too late...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Unknown008 View Post

    EDIT: A little too late...
    You have a different approach, so it's fine.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Quote Originally Posted by Jhevon View Post
    Given the ranges of \displaystyle \alpha \text{ and } \beta, clearly both \displaystyle \tan \alpha , ~ \tan \beta > 0, and so, their sum will be positive also. That is, \displaystyle 0 < \tan \alpha + \tan \beta.

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

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

    \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)..
    :-/
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Thank you so much!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Quote Originally Posted by Unknown008 View Post
    Introduce tan on both sides of the given equation:

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

    This becomes:

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

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

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

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

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

    Hence, we get

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

    EDIT: A little too late...
    Thank you...can you explain me how can you prove that " 1 - \tan\alpha\tan\beta is positive. and less than 1" ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by rectangle View Post
    Thank you...can you explain me how can you prove that " 1 - \tan\alpha\tan\beta is positive. and less than 1" ?
    He did explain. What didn't you get?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Oct 2010
    Posts
    11
    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...
    :-/
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by rectangle View Post
    Thank you...can you explain me how can you prove that " 1 - \tan\alpha\tan\beta is positive. and less than 1" ?
    Although you are given that both 0<\alpha<\frac{{\pi}}{2} and 0<\beta<\frac{{\pi}}{2}

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

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

    hence Tan\alpha<1 and 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
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Thank you so much.
    I am such an idiot.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    No, you are not.
    You persisted. That's how to learn. Keep it up.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    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:

    \tan \alpha + \tan \beta = \tan45

    What is true is

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

    Since \tan \alpha + \tan \beta is divided by something, it is not equal to 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove inequality
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 26th 2011, 02:49 AM
  2. Prove inequality
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: July 15th 2010, 04:28 AM
  3. prove an inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 4th 2009, 08:31 AM
  4. Prove this inequality
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 16th 2008, 06:08 AM
  5. prove inequality
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 11th 2007, 05:44 PM

Search Tags


/mathhelpforum @mathhelpforum