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Math Help - Difficulf exercise! I can't do please help me

  1. #1
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    Question Difficulf exercise! I can't do please help me

    Hello I can't do this exercise please help me
    Let a, b, c be positive real numbers such that abc=1. Prove that

    (a-1+1/b)(b-1+1/c)(c-1+1/a)lessthan or eqal1
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  2. #2
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    Quote Originally Posted by luckyvan
    Hello I can't do this exercise please help me
    Let a, b, c be positive real numbers such that abc=1. Prove that

    (a-1+1/b)(b-1+1/c)(c-1+1/a)lessthan or eqal1
     (a - 1 + \frac{1}{b})(b - 1 + \frac{1}{c})(c - 1 + \frac{1}{a})
     (\frac{{ab - b + 1}}{b})(\frac{{bc - c + 1}}{c})(\frac{{ac - a + 1}}{a})
     \frac{{(ab - b + 1)(bc - c + 1)(ac - a + 1)}}{{abc}}
     \frac{{(ab^2 c - abc + ab - b^2 c + bc - b + bc - c + 1)(ac - a + 1)}}{{abc}}
     \frac{{(ab - b^2 c + 2bc - c)(ac - a + 1)}}{{abc}}
     a + b + c - a^2 b - b^2 c - ac^2  + ab + bc + ac - 2
     ab(1 - a) + bc(1 - b) + ac(1 - c) + a + b + c - 2
    a(b - ab + 1) + b(c - bc + 1) + c(a - ac + 1) - 2

    Since a,b,c are all positive then
    a(b - ab + 1) + b(c - bc + 1) + c(a - ac + 1) must be lower or equal then 3.

    It needs proving but I think its easy to prove. I don't have time now but someone else can use this to prove my solution.
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  3. #3
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    I have tried to solve it a bit different.

    Putting abc instead of 1 (because 1=abc) we get
     (a - abc + ac)(b - abc + ab)(c - abc + bc)
    a(1 - bc + c)b(1 - ac + a)c(1 - ab + b)
    (1 - bc + c)(1 - ac + a)(1 - ab + b)

    Now, lets consider next cases:
    If a,b,c are all equal to 1 then (1 - bc + c)(1 - ac + a)(1 - ab + b) \le 1<br />
is true.

    If one of them is equal to 1 and others not (for example let be a = 1,bc = 1 )
    we would get:


    (1 - bc + c)(1 - ac + a)(1 - ab + b) =
    (1 - 1 + \frac{1}{b})(1 - \frac{1}{b} + 1)(1 - \frac{1}{c} + \frac{1}{c}) =
    \frac{1}{b}(2 - \frac{1}{b})

    Knowing that  c = \frac{1}{b} then \frac{1}{b}(2 - \frac{1}{b}) is equal to:
    c(2 - c) =
    2c - c^2

    Because c is positive, (2-c) can have only positive values 1 and 0 and for all other values (2-c) is always negative so
    2c - c^2  \le 1 is true.

    Hope this solve the problem.
    Last edited by DenMac21; February 16th 2006 at 04:40 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    It's a pretty little problem, isn't it? It's relatively easy to use the Calculus to show that a=b=1 (where c=1/ab) is an absolute maximum for the expression (and the max = 1 so the expression is less than or equal to 1), but I doubt that this is the approach that the problem was designed for.

    -Dan
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  5. #5
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    For what aproach is problem designed?

    What is the solution?
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