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

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    Inequality, not very hard but nice olympiad problem

    Find the smallest real number x, such that: For arbitrary real numbers a, b, c\geqslant x and a + b + c = 3 this inequality is correct: a^{3} + b^{3} + c^{3}\geqslant 3
    Last edited by Ununuquantium; September 1st 2007 at 08:51 AM.
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    Smile Smallest real number x where a,b,c >=x, a+b+c=3, a^3+b^3+c^3>=3

    Quote Originally Posted by Ununuquantium View Post
    Find the smallest real number x, such that: For arbitrary real numbers a, b, c\geqslant x and a + b + c = 3 this inequality is correct: a^{3} + b^{3} + c^{3}\geqslant 3
    Is the answer x = 0 ? (where a = b = c = 1)

    I got this by trial and error, but I am still half asleep right now.


    Mahurshi Akilla
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by Ununuquantium View Post
    Find the smallest real number x, such that: For arbitrary real numbers a, b, c\geqslant x and a + b + c = 3 this inequality is correct: a^{3} + b^{3} + c^{3}\geqslant 3
    I would say x = 1, but i'm not sure how to prove it mathematically.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mahurshi View Post
    Is the answer x = 0 ? (where a = b = c = 1)

    I got this by trial and error, but I am still half asleep right now.


    Mahurshi Akilla
    what about a = b = -1, c = 5

    then we could have x = -1

    Quote Originally Posted by janvdl View Post
    I would say x = 1, but i'm not sure how to prove it mathematically.
    what about a = b = 0 and c = 3

    then we could have x = 0


    Quote Originally Posted by Ununuquantium View Post
    Find the smallest real number x, such that: For arbitrary real numbers a, b, c\geqslant x and a + b + c = 3 this inequality is correct: a^{3} + b^{3} + c^{3}\geqslant 3
    this is a weird question. perhaps it should read "the greatest real number x," or "the smallest nonnegative real number x," since we can probably find an infinite number of combinations of numbers that fulfil the required conditions, especially since we don't have to restrict ourselves to integers (i can't prove this, but i think it makes sense). Consider even, a = \sqrt {2}, b = - \sqrt {2}, c = 3. those also fulfil the conditions given. so we could write x = - \sqrt {2}.

    In fact, (without loss of generallity) if a + b = 0, and c = 3, say, then we fulfil these conditions for an infinite set of numbers, since there are infinite possibilities for a and b such that a + b = 0. i think we can show that by induction. but let's just consider extremes here:

    what about a = \infty, b = - \infty, c = 3

    then we could say x = -\infty, so whatever x you tell me, i can tell you a smaller one for which it works

    (of course i realize that no number can equal infinity, and that \infty - \infty may not necessarilly be zero--or is it? but i just want to show we can get as small a number as we want for x)

    i could say x = -500000, then choose a = 500000, b = -500000 and c = 3, and those work!

    ok, enough rambling, i think you guys get my point, or am i not making sense at all?
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    I think x=-5 - but I am not sure and I cannot prove it...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ununuquantium View Post
    I think x=-5 - but I am not sure and I cannot prove it...
    did you read my post above? did you have a problem with it? are you sure you have the question right? as it stands, x can be anything i believe
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    No, the question is formed absolutely properly. Yor answer is wrong...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ununuquantium View Post
    No, the question is formed absolutely properly. Yor answer is wrong...
    you're sure the answer is around -5?
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    I am not absolutely sure, but I think that the answer is 5
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ununuquantium View Post
    Find the smallest real number x, such that: For arbitrary real numbers a, b, c\geqslant x and a + b + c = 3 this inequality is correct: a^{3} + b^{3} + c^{3}\geqslant 3
    Quote Originally Posted by Ununuquantium View Post
    I am not absolutely sure, but I think that the answer is 5
    ok. let's say you made a huge mistake, and the answer was actually even smaller, say -25.

    I claim the answer is x = -25

    To show this works, I choose a = -25, b = 25, c = 3

    is a,b,c >= x?
    yes! -25, 25 and 3 are all greater than or equal to -25. so the first condition is fulfilled.

    is a + b + c = 3? yes! -25 + 25 + 3 = 3. so our second condition is fulfilled.

    are a,b,c, and x real numbers? yes! they are integers. so that condition is fulfilled

    Now the moment of truth:

    is the inequality a^3 + b^3 + c^3 >= 3 correct?
    yes!

    (-25)^3 + (25)^3 + 3^3 = -(25)^3 + (25)^3 + 3^3 = 27 >= 3


    so all conditions hold for x = -25. so x = -25 is the answer
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    Ugh..... I cannot get through that problem...
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    Quote Originally Posted by Jhevon View Post
    ok. let's say you made a huge mistake, and the answer was actually even smaller, say -25.

    I claim the answer is x = -25

    To show this works, I choose a = -25, b = 25, c = 3

    is a,b,c >= x?
    yes! -25, 25 and 3 are all greater than or equal to -25. so the first condition is fulfilled.

    is a + b + c = 3? yes! -25 + 25 + 3 = 3. so our second condition is fulfilled.

    are a,b,c, and x real numbers? yes! they are integers. so that condition is fulfilled

    Now the moment of truth:

    is the inequality a^3 + b^3 + c^3 >= 3 correct?
    yes!

    (-25)^3 + (25)^3 + 3^3 = -(25)^3 + (25)^3 + 3^3 = 27 >= 3


    so all conditions hold for x = -25. so x = -25 is the answer
    Gee, Jhevon, are you sure it isn't x = -50?

    -Dan
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    Quote Originally Posted by Jhevon View Post
    ok. let's say you made a huge mistake, and the answer was actually even smaller, say -25.

    I claim the answer is x = -25

    To show this works, I choose a = -25, b = 25, c = 3

    is a,b,c >= x?
    yes! -25, 25 and 3 are all greater than or equal to -25. so the first condition is fulfilled.

    is a + b + c = 3? yes! -25 + 25 + 3 = 3. so our second condition is fulfilled.

    are a,b,c, and x real numbers? yes! they are integers. so that condition is fulfilled

    Now the moment of truth:

    is the inequality a^3 + b^3 + c^3 >= 3 correct?
    yes!

    (-25)^3 + (25)^3 + 3^3 = -(25)^3 + (25)^3 + 3^3 = 27 >= 3


    so all conditions hold for x = -25. so x = -25 is the answer
    you're right. x can be a number that's much smaller too. i didn't realize that when i posted earlier.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Gee, Jhevon, are you sure it isn't x = -50?

    -Dan
    of course i was being sarcastic here

    as i said, any x you choose, i choose a smaller x and still find real numbers a,b, and c that fulfill all conditions
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  15. #15
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    I think that, "for arbitrary" in the contents means "for any(every)"
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