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Triangle Problem

  1. Sep 29, 2012 #1

    FeDeX_LaTeX

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    1. The problem statement, all variables and given/known data

    Let a, b and c be the lengths of the sides of a triangle. Suppose that ab + bc + ca = 1. Show that (a+1)(b+1)(c+1) < 4.


    2. Relevant equations
    N/A (Olympiad problem)


    3. The attempt at a solution
    I've made a few inroads at this problem, but whether or not they're going to take me in the right direction, I'm not sure. I used the arithmetic/harmonic mean inequality to show that:

    [tex]9abc \leq a + b + c[/tex]

    and I've used substitutions from their original equation to arrive at

    [tex]8(a+b+c) \leq (a+b)(1+9c^2) + (b+c)(1+9a^2) + (a+c)(1+9b^2)[/tex]

    but I'm puzzled at where to go from here. I don't think it would be useful to consider specific cases such as the equilateral triangle (a = b = c), since we appear to be trying to prove a general case, and I know I mustn't start from their solution and show that it's equal to the original equation.

    Am I taking the right approach? Is a geometric approach better? Any hints/tips anyone can give, without giving me the answer?

    Thanks.
     
  2. jcsd
  3. Sep 29, 2012 #2

    FeDeX_LaTeX

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    I recognise that

    [tex](a + 1)(b + 1)(c + 1) - 2(ab + bc + ca) - 2 = (a - 1)(b - 1)(c - 1)[/tex]

    and so

    [tex](a + 1)(b + 1)(c + 1) - 4(ab + bc + ca) = (a - 1)(b - 1)(c - 1)[/tex]

    ...I feel like I'm close, but not sure where to place the final step.

    EDIT: Ugh, it's obvious... clearly the RHS is smaller than zero, so...

    [tex](a+1)(b+1)(c+1) < 4(ab + bc + ca)[/tex]
    [tex](a+1)(b+1)(c+1) < 4[/tex]

    Done.
     
    Last edited: Sep 29, 2012
  4. Sep 30, 2012 #3

    FeDeX_LaTeX

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    How do I show that a, b and c are all smaller than 1 though? It's a requirement for the above. I thought it could just be assumed but it is not as simple as that.
     
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