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Prove or disprove this inequality

  1. Oct 11, 2004 #1
    :confused:
    [tex]a^2 - b^2 - c^2 - d^2 > 0[/tex]
    [tex]A^2 - B^2 - C^2 - D^2 = 1[/tex]
    [tex]A > 1[/tex]
    [tex]a > 0[/tex]
    Prove or disprove [tex]Aa + Bb + Cc + Dd > 0[/tex]

    after 3 days of trying I give up
    can anyone give a clue?
     
  2. jcsd
  3. Oct 11, 2004 #2
    what I have done:
    I used [tex](a + b)^2 = a^2 + b^2 + 2ab[/tex], but it leads to nothing.
    I have tried Lagrange Multipliers, but the only extremum is a saddle point.
     
  4. Oct 11, 2004 #3
    trying to disprove it
    I found no counter-example.
     
  5. Oct 11, 2004 #4
    (a+A)^2 + (b+B)^2 + (c+C)^2 + (d+D)^2 > 0

    -- AI
     
  6. Oct 11, 2004 #5
    I can't see.
    then?
     
  7. Oct 11, 2004 #6
    Are b,c,d >0 and B,C,D>1 ?? if so, no need to prove anything since it is obvious.
    If not, I think we can't say the given formula >0 or <0
    You can try subsitute some values of b,c,d to guess a, then values for B,C,D to calculate A, you will see it.

    Am I incorrect ?
     
    Last edited: Oct 11, 2004
  8. Oct 11, 2004 #7
    Let me try to give some motivation about the inequaility.

    Consider the equivalent problem in 2D. That is,
    [tex]a^{2}-b^{2}>0, a>1[/tex]
    [tex]A^2-B^2=1, A>1[/tex]
    Prove or disprove [tex]aA+bB>0[/tex]
    In 2D, the region represented by the first equation is like a "quadrant" rotated by 45 degrees. (Try to plot it.) The set represented by the second equation is one component of a hyperbola included in the first region. Note that Aa+Bb is just the usual *dot product* between vector (a, b) and (A, B). Can you see why the inequality holds in the case?

    Try to do the problem for 3D and then generalise it.
     
  9. Oct 11, 2004 #8
    Wow, my gal bit me .:redface:
     
  10. Oct 12, 2004 #9
    To Wong,

    The angle between the vectors is always less than 90?
    I can see it in 2D and 3D, but not in 4D, but how to write a proof?
     
  11. Oct 12, 2004 #10
    hi kakarukeys.

    To prove it, first note that the set represented by [tex]a^{2}-b^{2}-b^{2}-c^{2}=1, a>1[/tex] is in fact a subset of [tex]A^2-B^2-C^2-D^2>0, A>0[/tex]. Then what we need to prove becomes
    Prove [tex]Aa+Bb+Cc+Dd>0[/tex], where (A, B, C, D) and (a, b, c, d) both satisfies,
    [tex]h^2-i^2-j^2-k^2>0, h>0[/tex]
    [tex]A^2>B^2+C^2+D^2[/tex]
    [tex]A>(B^2+C^2+D^2)^{\frac{1}{2}}[/tex]
    Similarly,
    [tex]a>(b^2+c^2+d^2)^{\frac{1}{2}}[/tex]
    Put those expression into Aa+Bb+Cc+Dd, does it remind you of some inequality?
    This can be generalised to any dimensions.

    Somtimes it can be quite difficult to think of a proof for inequalities, even though it is quite trivial geometrically.
     
  12. Oct 14, 2004 #11
    Intuitive guide, I have spotted the 'subset' clue.
    but I still can't see the solution.

    [tex]Aa > \sqrt{B^2 + C^2 + D^2}\sqrt{b^2 + c^2 + d^2} \geq Bb + Cc + Dd[/tex]
    (Cauchy-Schwarz inequality)

    And so [tex]Aa - Bb - Cc - Dd > 0[/tex]
    that's not we want.

    Note [tex]Bb + Cc + Dd[/tex] can be negative.
     
  13. Oct 14, 2004 #12
    ok COOL
    [tex]Aa > \sqrt{B^2 + C^2 + D^2}\sqrt{b^2 + c^2 + d^2} = \sqrt{(-B)^2 + (-C)^2 + (-D)^2}\sqrt{b^2 + c^2 + d^2}
    \geq - Bb - Cc - Dd[/tex]
     
  14. Oct 14, 2004 #13
    Thumbs up!
     
  15. Oct 14, 2004 #14
    Interesting indeed!!!!
    I was lost with my initial thoughts !!!!
    Bravo!!

    -- AI
     
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