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Math Help - triple inequalities

  1. #1
    shosho
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    triple inequalities

    i really dontknow this questoin any help would be great

    prove that if real numbers a,b,c satisfy

    a + b + c>0, ab +ac + bc>0 , abc>0

    then each of a,b,c is positive


    thank you in advance
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shosho View Post
    i really dontknow this questoin any help would be great

    prove that if real numbers a,b,c satisfy

    a + b + c>0, ab +ac + bc>0 , abc>0

    then each of a,b,c is positive


    thank you in advance
    It ain't elegant but it works:

    ab + ac + bc > 0

    a(b + c) + bc > 0

    1) b + c > -\frac{bc}{a} if a > 0
    or
    2) b + c < -\frac{bc}{a} if a < 0
    or
    3) bc > 0 if a = 0

    Now:
    3) is the easiest case. abc > 0 and a = 0 are contradictory. Thus a \neq 0.

    So consider case 1):
    1) b + c > -\frac{bc}{a} if a > 0

    We know that abc > 0 and a is positive, so bc > 0. This means that either both b, c are positive or both b, c are negative.

    If both b, c are positive then b + c > -\frac{bc}{a} is trivial since the RHS is negative, and we have no contradiction.

    If both b, c are negative then
    b + c > -\frac{bc}{a}

    \frac{bc}{a} + c > -b

    c \cdot \left ( \frac{b}{a} + 1 \right ) > -b

    c \cdot \left ( \frac{a + b}{a} \right ) > -b

    Now if a + b > 0 then
    c > \left ( \frac{a}{a + b} \right ) \cdot -b
    is a contradiction because c < 0 and the RHS is negative.

    If a + b < 0 and we know that c is negative then a + b + c > 0 cannot be true because (a + b) + c < 0, a contradiction.

    Thus we are left with a + b = 0.
    Thus
    c \cdot \left ( \frac{a + b}{a} \right ) > -b

    c > 0
    another contradiction.

    Thus both b and c cannot be negative.

    Thus case 1) says that if a is positive, so are both b and c.

    Case 2) works in a similar fashion to show that, at best, all three of a, b, and c must be negative. (I leave this for you to show.) This again contradicts a + b + c > 0.

    -Dan
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  3. #3
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    hey shosho seems to understand this problem.. but im not quite sure i do.
    i dont understand the contradiction bit.. could you explain more? it seems like a reeli hard question
    thanks
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ieatfood View Post
    hey shosho seems to understand this problem.. but im not quite sure i do.
    i dont understand the contradiction bit.. could you explain more? it seems like a reeli hard question
    thanks
    The basic idea behind what I was doing is this:
    Given a real number a, we know that only one of the following possibilities is true:
    a < 0
    a = 0
    a > 0

    I used this idea in two places in the proof of case 1): First in showing that if a is positive then b,c must both be either positive or negative. I showed that there is no contradiction if both b, c are positive. Then I showed that if b,c are both negative that a contradiction exists by looking at a + b in reference to >,=,<. There was a contradiction in all three cases, so I showed that a + b, a real number, is neither >, =, or < 0, an impossibility.

    -Dan
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