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Math Help - Inequality, not very hard but nice olympiad problem

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by terafull View Post
    I think that, "for arbitrary" in the contents means "for any(every)"
    i did use arbitrary numbers. infinitely many arbitrary numbers, in fact.

    something is wrong with the question.
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  2. #17
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    Quote Originally Posted by Jhevon View Post
    i did use arbitrary numbers. infinitely many arbitrary numbers, in fact.

    something is wrong with the question.
    No, you are wrong. You said, you can find numbers a,b,c which fulfil conditions for any x, but you must find x, which for any possible a,b,c (a+b+c=3) is filfuled. for example you said x=-50. it is wrong but I can find numbers a,b,c which isnt filfuled (a^3+b^3+c^3>=3).. for example a=-50 b=26 c=27
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by terafull View Post
    No, you are wrong. You said, you can find numbers a,b,c which fulfil conditions for any x, but you must find x, which for any possible a,b,c (a+b+c=3) is filfuled. for example you said x=-50. it is wrong but I can find numbers a,b,c which isnt filfuled (a^3+b^3+c^3>=3).. for example a=-50 b=26 c=27
    it says for arbitrary values THAT FULFILL THE GIVEN CONDITIONS. you said yourself that arbitrary means "any." if it means "any" why does it matter if i pick them or not? i found arbitrary values that fulfill all conditions, but for everyone i find, i can find smaller and smaller x's, thus there is no minimum value for the x.

    in math, to disprove something, all you need is ONE counter example, and you are allowed to choose it if you can find one. here i found infinitely many counter-examples. it is easy to satisfy the equations and inequalities, but there is no minimum for the x value as the problem claimed. since i can find arbitrary a,b, and c values that fulfill all requirements, but require a smaller x value than any that i chose before. if you look at my post, you will realize that i could choose the a,b,c first and then find the x. a,b, and c are arbitrary, so i can choose any a,b,c i want, as long as they fulfill the required conditions. (your example does not fulfill the conditions by the way).
    Last edited by Jhevon; October 16th 2007 at 09:29 AM.
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  4. #19
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    my example doesnt fulfil a^3+b^3+c^3>=3 because chosen x is wrong... you must first find x and then for every posibility a,b,c which fulfil a+b+c=3, this a^3+b^3+c^3>=3 is true... I repeat fo every combination of a,b,c , not one I found this ONE exapmle to disprove your chosen x (-50) for example x=0, can you find a,b,c, which doesnt filful equation and inequality?
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by terafull View Post
    my example doesnt fulfil a^3+b^3+c^3>=3 because chosen x is wrong... you must first find x and then for every posibility a,b,c which fulfil a+b+c=3, this a^3+b^3+c^3>=3 is true... I repeat fo every combination of a,b,c , not one I found this ONE exapmle to disprove your chosen x (-50) for example x=0, can you find a,b,c, which doesnt filful equation and inequality?
    ah, i see what you are saying! and after reading the question again, i believe you are correct, my bad . i do not think x = 0 is the answer though. what do you think the answer is?
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  6. #21
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    I dont know, but I think x is smaller than 0. -1 is true in my opinion. I tried go down with x (x=a) and b=c (I think it is the worst case), I found that x=-5. a=-5 b=c=4
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  7. #22
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    x=-5 for sure. This is not hard question, but I cannot get through it. I know the answer, because it is given in a book - but there is no solution . Maybe anyone can find any good, full soluton/proof??
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