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Can someone prove this?

  1. Apr 4, 2003 #1
    my teacher gave me a question in the class...

    then prove:
    (1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002)

    can someone help me to prove it?

    kian yew.
  2. jcsd
  3. Apr 4, 2003 #2


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    What conditions are put on a, b, c?

    It's fairly easy to prove that (1/a)+(1/b)+(1/c)=1/(a+b+c)
    is never true for a, b, c positive and that, therefore,

    (1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002)

    is never true for a, b, c real.
  4. Apr 4, 2003 #3
    she just gave me this. why it's never true for a,b,c is real?
    can't a,b,c be any real numbers
  5. Apr 4, 2003 #4


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    In general 1/a+1/b+1/c=1/(a+b+c) is not generally true. I am assuming that it is supposed to be a given that it is true for this problem. Is this so?

    If so, b=a and c=-a is a workable solution set for the first assumption. However, it is a counter-example for the second equation. The second equation is disproved, not proved.

  6. Apr 4, 2003 #5


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    Actually, the general solution for 1/a+1/b+1/c=1/(a+b+c) is:

    a=-b or a=-c or b=-c with no constraint on the other variable.

    If the second equation used odd powers instead of even powers, it would be true, given the first equation.

  7. Apr 4, 2003 #6
    Take a+b=0 and let c free. Then the second is satisfied because its trivial. These are indeed the only possible solution types which can be applied to this.
  8. Apr 4, 2003 #7


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    The second equation is not satisfied, it is disproved. if a+b=0, a=-b.

    Then, when each is raised to an even power like 2002, the sign change disappears. An odd power would make it work.

  9. Apr 4, 2003 #8
    Oh ****! I read 1/(x+y+z)^2002

    You're right. Momentaneous offuscation :frown:
  10. Apr 4, 2003 #9


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    (1/a)+(1/b)+(1/c)= (ab+ ac+ bc)/abc so
    if (1/a)+(1/b)+(1/c)=1/(a+b+c) we have
    (ab+ac+bc)/abc= 1/(a+b+c)

    Now multiply both sides of the equation by abc and a+b+c to "clear the fractions":

    (a+b+c)(ab+ac+bc)= abc
    Multiply the left side out:
    a2b+ a2c+ abc+ ab2+abc+b2c+ abc+ ac2+ bc2= abc.

    One of those 3 "abc"s on the left cancels the one on the right so
    ab2+b2c+ ac2+ bc2+ 2abc= 0. If a, b, c are all real, this is impossible.

    In particular, in (1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002) all the denominators are to even powers and so are positive as long as a, b, are real.
  11. Apr 4, 2003 #10
    Okay I followed you to here, but can't see why that is impossible for all reals. Maybe I'm missing something, could you help me out.
  12. Apr 4, 2003 #11


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    Indeed, that seems to me to be wrong. (Though I might just be mistaken). Can't you rewrite that as 0 = (b + c)(ab+ac+bc) ?
    Then clearly if b+c = 0, the equation is correct...
  13. Apr 4, 2003 #12


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    requires a, b, and c and a+b+c to be non-zero for the divisions to be defined. Said that, we can proceed by passing to a common denominator on the left side

    (ab+ac+bc)/abc= 1/(a+b+c)

    Now multiply both sides of the equation by abc and a+b+c to "clear the fractions" which is possible only in presence of the previous conditions!!!

    (a+b+c)(ab+ac+bc)= abc

    Multiply out the left side

    a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2= abc.


    a^2b+a^2c+ab^2+2abc+b^2c+ac^2+bc^2+2abc= 0

    collect c from all the terms where it appears with power 1

    a^2b+ab^2+ac^2+bc^2+(a^2+2ab+b^2)c= 0

    recognize square of binomial and collect c^2 terms and ab from the two remaining ones

    ab(a+b)+(a+b)c^2+(a+b)^2c= 0

    collect (a+b)

    (a+b)[ c^2+(a+b)c+ab ]= 0

    It easy to see if you that the second term can be rewritten as a product of two binomials and have

    (a+b)(b+c)(c+a)= 0

    That requires that at least one (but not all three of them, otherwise all should be zeros against our hypotheses) of the following holds

    a=-b aut b=-c aut c=-a

    For(1/a^2002)+(1/b^2002)+(1/c^2002)=1/(a^2002 + b^2002 + c^2002) all you need to do is to substitute a->a^2002, etc. in this case by the way the previous formulas cannot hold for any number different from zero and hence the equality is always impossible.

    Hope this helped
  14. Apr 4, 2003 #13
    thanks to all your help, but i'm thinking another way to do it and i don't know how to become this:

    first, turn (1/a)+(1/b)+(1/c)=1/(a+b+c) into
    (1/a^2)+(1/b^2)+(1/c^2)=1/(a^2+b^2+c^2) then turn again into (1/a^3)+(1/b^3)+(1/c^3)=1/(a^3+b^3+c^3) and then.....
    from this i can say that (1/a^n)+(1/b^n)+(1/c^n)=1/(a^n+b^n+c^n). the problem for me is i don't know how to make the first few line that i show above. or this way is wrong? can i use induction method?

    kian yew.
  15. Apr 5, 2003 #14


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    Yes it is wrong!
    please check my previous answer!

    induction here is a useless complication all you need to consider is that is an even power

  16. Apr 6, 2003 #15
    you wrong
    i think this cant use induction
    because when n=1, this equation also cant do
    so i think maybe a=b=c=1 .....:)
  17. Apr 6, 2003 #16
    I guess it doesn't work for anything.
    Last edited: Apr 7, 2003
  18. Apr 7, 2003 #17

    mistake...a,b,c cant be 1....
    because the 1/1 + 1/1 + 1/1 cant be 1/(1+1+1)
    haha....so i think this equation are no solve......
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