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Math Help - Hard/beautiful inequality --help!!

  1. #1
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    Smile Hard/beautiful inequality --help!!

    Pls help to prove this inequality which I think is quite neat:

    --------------
    3 (a/b + b/a + b/c + c/b + a/c + c/a) + (1 + a) (1 + b) (1 + c) (c/b + c/a) (b/a + b/c) (a/b + a/c)
    >=
    6 abc + 6 + 9(ab + bc + ac + a + b + c) + 3(ab/c + bc/a + ac/b)
    --------------

    with a, b, c are positive reals

    If it helps, equality occurs when a=b=c=2. Also, can any one helps to prove (2, 2, 2) is the only point at which equality occur.

    Thanks
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  2. #2
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    Very nice problem! Could you tell me the source of this problem? Thanks.
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  3. #3
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    I discovered this inequality while working on a mathematical modelling of a genetic system with negative feedback loops. I'm working in systems biology area.

    I know for sure the inequality holds and can prove it computationally. I'm looking for a rigorous, mathematical proof though. Have you tried it out? Do you have any ideas on tackling it?

    Cheers,
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  4. #4
    Junior Member Serena's Girl's Avatar
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    Always greater than or equal to 2

    I am not certain if this will be helpful, but I'm starting out with the observation that for a pair of positive reals A and B:

    <br />
\frac {A} {B} + \frac {B} {A} \geq 2<br />

    This inequality can be proven simply by writing it as

    <br />
\frac {A^2 + B^2} {AB} \geq \frac {2AB} {AB}<br />

    and then

    <br />
\frac {A^2 + B^2 - 2AB + 2AB} {AB} \geq \frac {2AB} {AB}<br />

    <br />
\frac {(A - B)^2 + 2AB} {AB} \geq \frac {2AB} {AB}<br />

    Since (A - B)^2 will always be at least zero, the inequality holds true.

    Hope that helps.
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  5. #5
    Junior Member Serena's Girl's Avatar
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    Always greater than or equal to 3?

    I am now attempting to prove that

    <br />
\frac {ab} {c} + \frac {bc} {a} + \frac {ac} {b} \geq 3<br />

    It seems to me it is true, but I haven't found a mathematical proof for it yet. This is as far as I've gone:

    <br />
\frac {a^2 b^2 + b^2 c^2 + a^2 c^2} {abc} \geq \frac {3abc} {abc}<br />

    Simplifying to:

    <br />
\frac {ab(ab - c) + bc(bc - a) + ac(ac - b) + 3abc} {abc} \geq \frac {3abc} {abc}<br />

    So it's now a matter of proving that ab(ab - c) + bc(bc - a) + ac(ac - b) is at least zero. Any takers?
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