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Posers and Puzzles

Posers and Puzzles

  1. 03 Mar '08 16:43
    a < b < c < d

    prove:
    (a + b + c + d)^2 > 8 (ac + bd)
  2. Subscriber coquette On Vacation
    Already mated
    03 Mar '08 21:35
    no
  3. 03 Mar '08 23:08
    Originally posted by coquette
    no
    no? it's wrong?
  4. 04 Mar '08 16:50
    The rearrangement inequality. cbf working it out though
  5. Standard member wolfgang59
    Infidel
    04 Mar '08 20:53
    Originally posted by serigado
    a < b < c < d

    prove:
    (a + b + c + d)^2 > 8 (ac + bd)
    Consider
    {(a+b)-(c+d)}^2 > 0

    Expand
    (a+b)^2 + (c+d)^2 - 2(a+b)(c+d) > 0

    * Therefore
    (a+b)^2 + (c+d)^2 > 2(a+b)(c+d)


    Now
    {a+b+c+d}^2 = {(a+b)+(c+d)}^2
    = (a+b)^2 + (c+d)^2 + 2(a+b)(c+d)

    Now substituting for (a+b)^2 + (c+d)^2 from * gives;

    {a+b+c+d}^2 > 4(a+b)(c+d)
    expand

    {a+b+c+d}^2 > 4(ac + bc + ad + bd)
    {a+b+c+d}^2 > 4(ac + bc) + 4(ad + bd)

    Let b=a+x (where x is =ve since we know b>a)

    Substitute
    {a+b+c+d}^2 > 4ac + 4(a+x)c + 4(b-x)d + 4bd

    {a+b+c+d}^2 > 8ac + 4xc + 8bd -4xd

    {a+b+c+d}^2 > 8(ac + bd) + 4x(c-d)

    Therefore
    {a+b+c+d}^2 > 8(ac + bd)
  6. Standard member wolfgang59
    Infidel
    04 Mar '08 21:03
    Just one flaw with my proof.
    Its wrong!
  7. 06 Mar '08 06:27 / 1 edit
    Originally posted by wolfgang59
    {a+b+c+d}^2 > 8(ac + bd) + 4x(c-d)
    as d>c, c-d < 0 so next step is flawed
  8. 06 Mar '08 06:57
    my solution, based on wolfgang:

    Consider
    {(a+d)-(b+c)}^2 > 0

    Expand
    (a+d)^2 + (b+c)^2 - 2(a+d)(b+c) > 0

    * Therefore
    (a+d)^2 + (b+c)^2 > 2(a+d)(b+c)


    Now
    {a+b+c+d}^2 = {(a+d)+(b+c)}^2
    = (a+d)^2 + (b+c)^2 + 2(a+d)(b+c)

    Now substituting for (a+d)^2 + (b+c)^2 from * gives;

    {a+b+c+d}^2 > 4(a+d)(b+c)
    expand

    {a+b+c+d}^2 > 4(ab + ac + bc + bd)

    Consider - ab + ac - cd + bd
    = b(d-a) - c(d-a)
    = (b-c)(d-a)
    4(ab + ac + bc + bd):
    {a+b+c+d}^2 > 4(ab + ac + bc + bd - ab + ac - cd + bd)
    {a+b+c+d}^2 > 4(2ac + 2bd)

    Hence:
    {a+b+c+d}^2 > 8(ac + bd)
  9. 06 Mar '08 13:03 / 5 edits
    Consider - ab + ac - cd + bd
    = b(d-a) - c(d-a)
    = (b-c)(d-a)


    You only forgot to say this quantity is negative because a < b< c< d


    4(ab + ac + bc + bd):
    {a+b+c+d}^2 > 4(ab + ac + bc + bd - ab + ac - cd + bd)
    {a+b+c+d}^2 > 4(2ac + 2bd)


    And forgot to mention bc - cd is negative too, so the inequality is maintained
  10. 06 Mar '08 14:27 / 1 edit
  11. 06 Mar '08 14:55
    Originally posted by sb3700
    my solution, based on wolfgang:

    Consider
    {(a+d)-(b+c)}^2 > 0

    Expand
    (a+d)^2 + (b+c)^2 - 2(a+d)(b+c) > 0

    * Therefore
    (a+d)^2 + (b+c)^2 > 2(a+d)(b+c)


    Now
    {a+b+c+d}^2 = {(a+d)+(b+c)}^2
    = (a+d)^2 + (b+c)^2 + 2(a+d)(b+c)

    Now substituting for (a+d)^2 + (b+c)^2 from * gives;

    {a+b+c+d}^2 > 4(a+d)(b+c)
    expand

    {a+b+c+d}^2 > 4(ab ...[text shortened]... c + bc + bd - ab + ac - cd + bd)
    {a+b+c+d}^2 > 4(2ac + 2bd)

    Hence:
    {a+b+c+d}^2 > 8(ac + bd)
    {a+b+c+d}^2 > 4(a+d)(b+c)
    expand

    This far I understand, but it gets confusing from here on. Where does the bc term come from in the next line?

    {a+b+c+d}^2 > 4(ab + ac + bc + bd)

    Consider - ab + ac - cd + bd
    = b(d-a) - c(d-a)
    = (b-c)(d-a)
  12. 06 Mar '08 16:01
    I get it now. Your proof looks good apart from the mix up with bc and cd.