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Prove ( a + b + c )^2 <= 4( ab + bc + ca )

  1. Nov 10, 2006 #1
    The Problem;
    If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

    Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

    I haven’t make much progress. I’ve been trying;

    ( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

    But I haven’t been able to figure out a way to get the right side manipulated. I tried a couple of substitutions, but couldn’t justify a<b or such to be able to use the hint.

    Can anyone give me a push in the right direction?

    Thanks,
    Bernie
     
    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 10, 2006 #2

    acm

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    I would start the question by expanding the RHS of the equasion to:
    a^2 + b^2 + c^2 + 2(ab + bc + ac)

    Now, It is obvious that (a + b + c)^2 >= 2(ab + bc + ac)
    Hence 2(a + b + c)^2 >= 4(ab + bc + ac)
    Therefore we can assume that 4(ab + bc + ac) = 2(a + b + c)^2
    And as (a + b + c)^2 <= 2(a + b + c)^2
    Therefore (a + b + c)^2 <= 4(ab + bc + ac)
     
  4. Nov 11, 2006 #3
    ACM,

    Thanks for the quick reply. I think there is a flaw in your logic. The statements workout to;

    2(a + b + c)^2 > (a + b + c)^2 > 2(ab + bc + ac)
    2(a + b + c)^2 > 4(ab + bc + ac) > 2(ab + bc + ac)

    Therefore 4(ab + bc + ac) > (a + b + c)^2

    Which is saying that:
    x > y
    x > z
    Therefore y > z

    Which is not always true. It is true sometimes but not for all conditions.

    Is there a way to prove that because a + b > c, b + c > a, c + a > b then the above logic is always true?

    Bernie
     
  5. Nov 11, 2006 #4

    StatusX

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    Expand out the square and cancel like terms. Then use their clue.
     
  6. Nov 11, 2006 #5

    acm

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    No problem, Should have though that through.
     
  7. Nov 11, 2006 #6

    matt grime

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    Either a=b and you can use that fact, or either a<b, or b<a and wlog we may assume that a<b in this case. (or, they're either all equal, when it is trivially true, or we may assume a<b without loss of generality - the labels of a,b,c are obviously not important).
     
  8. Nov 11, 2006 #7
    StatusX,

    Cancel out which like terms? Expanding out the square gives;
    ( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

    Which of these terms can be canceled?

    Bernie
     
  9. Nov 11, 2006 #8

    StatusX

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    With the other side.
     
  10. Nov 11, 2006 #9
    And then rewrite 2ab as ab+ab, etc. Also, I don't think equality ever holds, unless you meant a + b >= c, b + c >= a, c + a >= b.
     
  11. Nov 12, 2006 #10
    OK, here we go;

    Prove:
    If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).

    Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.

    (a+b+c)^2 = (a+b+c)^2 (Reflective property of equality)

    (a+b+c)^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

    a + b > c
    c(a+b) > c^2 fact of inequity
    ca + cb > c^2

    b + c > a
    a(b+c) > a^2 fact of inequity
    ab + ac > a^2

    c + a > b
    b(c+a) > b^2 fact of inequity
    bc + ab > b^2

    From above
    (a+b+c)^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca

    By substitution
    (a+b+c)^2 < (ab + ac) + (ca + cb) + (bc + ab) + 2ab + 2bc + 2ca

    (a+b+c)^2 < 2ab + 2ac + 2bc + 2ab + 2bc + 2ca
    (a+b+c)^2 < 4ab + 4ac + 4bc
    (a+b+c)^2 < 4(ab + ac + bc)

    (a+b+c)^2 <= 4(ab + ac + bc) ???????????????????????

    Can I make this last step from < to <=? If so is there a principal or property that justifies it?

    Thanks,
    Bernie
     
  12. Nov 12, 2006 #11

    radou

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    Think about what happens if a = b = c = 0.
     
  13. Nov 12, 2006 #12

    StatusX

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    If something is less than something else, it is certainly also less than or equal to it. It's probably a typo though, and you were supposed to prove your next to last line.
     
  14. Nov 12, 2006 #13
    The proof is definately <=, but I think I found the justification. I don't have my notes handy, but I think it's disjunctive addtion justifies that if p then (p or q). Also if (p or q) and (~q) then p.

    Bernie
     
  15. Nov 12, 2006 #14

    StatusX

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    Yea, although it's pretty obvious if you just think about it. Also, I'm not saying you copied the question wrong, I'm saying it was wrong in the book. It's not strictly incorrect, but it's completely unnatural to use a [itex]\leq[/itex] in place of a < when equality is not possible.
     
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