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Prove a^2+b^2+c^2

  1. Aug 5, 2004 #1


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    can any one prove this :
    a^2+b^2+c^2 ( is greater or equal to ) ab + ac + bc
  2. jcsd
  3. Aug 5, 2004 #2


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    Dearly Missed

    A very important identity in mathematics is:
    Note that this is readily transformed into another important identity:
    now, try to rewrite
  4. Aug 5, 2004 #3
    This is like....
    a^2+b^2 >= ab, right?

    take (a-b)^2 >= 0.
    so a^2+b^2 >= 2ab >=ab.

    so now u want a^2+b^2+c^2,
    so take (a-b-c)^2, and (a+b-c)^2, etc...
    and do the same sort of thing, a bit trickier though.
  5. Aug 5, 2004 #4


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    arildno's approach is much nicer...but forgive his efforts at humor.
  6. Aug 5, 2004 #5

    I miss the dinosaurs. -ss
  7. Aug 5, 2004 #6
    How about...we assume a^2 + b^2 + c^2 < ab + ac + bc and consider a special case where 'a' being minus and absolute value of 'a' is greater than the absolute value of b for ovious reason to lead a contradiction. So it will prove the negation of what assumed is true.
  8. Aug 6, 2004 #7

    matt grime

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    it will only provide a contradiction in the special case where a is negative and in abs value greater then b, which need not be true.
  9. Aug 7, 2004 #8
    Use the identity for [tex]a^2 + b^2 + c^2 - ab - bc - ca[/tex]:

    a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^{2} + (b-c)^{2} + (c-a)^{2}]

    The right hand side is always greater than or equal to zero (equality in the case a = b = c). This proves the result.

    Hope that helps...

    Last edited: Aug 7, 2004
  10. Aug 7, 2004 #9


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    This is exactly what arildno was saying - without actually putting the spoon in the mouth.
  11. Aug 7, 2004 #10
    Oh well I didn't quite figure that out and since the question seemed unanswered to me so I went ahead and posted the solution (put the spoon in the mouth if you like it that way) :-). Its been quite a while since it was posted anyway.

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