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Math Help - Inequality help

  1. #1
    Newbie
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    May 2007
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    Inequality help

    I need help with the following inequality:

    Show that   (a^2 - b^2)(c^2- d^2) \leq (ac-bd)^2 for all  a,b,c,d

    Any help will be appreciated. Thank you.
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  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hi dude!

    Quote Originally Posted by MagicS06 View Post
    I need help with the following inequality:

    Show that   (a^2 - b^2)(c^2- d^2) \leq (ac-bd)^2 for all  a,b,c,d

    Any help will be appreciated. Thank you.
     (a^2 - b^2)(c^2- d^2) \leq (ac-bd)^2

    <=>
    [
    you know what I did? First I solved
     (a^2 - b^2)(c^2- d^2) =  - a^2d^2 + a^2c^2 + b^2d^2 - b^2c^2
    and then I solved
    (ac-bd)^2 = a^2c^2 - 2abcd + b^2d^2
    ]

     - a^2d^2 + a^2c^2 + b^2d^2 - b^2c^2 \le a^2c^2 - 2abcd + b^2d^2

    step: -a^2c^2

    - a^2d^2 + b^2d^2 - b^2c^2 \le - 2abcd + b^2d^2

    step: -b^2d^2

    - a^2d^2 - b^2c^2 \le - 2abcd

    step: multiplied by (-1)

    a^2d^2 + b^2c^2 \ge 2abcd

    We know

    (ad-bc)^2 \ge 0 (It is obvious, do you agree?)

    and therefor it is

    (ad-bc)^2 = a^2d^2 -2adbc + b^2c^2 \ge 0

    step: +2adbc (this is equal to 2abcd)

    => a^2d^2 + b^2c^2 \ge 2adbc

    q.e.d.

    Any questions?

    Best regards, Rapha
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