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Putnam and beyond prob 124

  1. May 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Let a,b,c be the side lengths of a triangle with semiperimeter 1. Prove that

    [tex]1 < ab + bc + ca - abc \leq 28/27[/tex]


    2. Relevant equations
    the AM-GM inequality:

    If x_1,...,x_n are nonnegative real numbers, then

    [tex]\frac{\sum x_i}{n} \leq \left( \Pi x_i\right)^{1/n}[/tex]


    3. The attempt at a solution
    I can get "close" to the second inequality. Applying AM-GM to a+b-c,a-b+c,-a+b+c gives

    [tex]8/27 \geq a^3 +b^3 +c^3 -2abc +a^2b+ab^2+ac^2 +a^2 c+bc^2 +b^2 c [/tex]

    Applying AM-GM to a,b,c gives

    [tex]8/27 \geq abc [/tex]

    and then we can add the two inequalities to obtain


    [tex]16/27 \geq a^3 +b^3 +c^3 -abc +a^2b+ab^2+ac^2 +a^2 c+bc^2 +b^2 c [/tex]

    Ok maybe thats not really close...
     
    Last edited: May 16, 2008
  2. jcsd
  3. May 16, 2008 #2
    I would probably just look at the limiting cases.

    First, arbitrarily set a <= b <= c.

    What if one of the sides becomes arbitrarily small? Well, if a =~ 0, then b = c and
    ab + bc + ca - abc = 0 + 1 + 0 - 0 = 1.

    Since a triangle with sides of length 0, 1, 1 isn't technically a triangle, we don't include this lower limiting number.

    Next, we guess the only other special triangle we can think of: the equilateral triangle. Thus a = b = c = 2/3, and ab + bc + ca - abc = 4/9 + 4/9 + 4/9 - 8/27 = 28/27. Since an equilateral triangle is a triangle, this is acceptable, and we keep the equality.

    Can this reasoning be made mathematically precise? Let's see.

    Let's call ab + bc + ca - abc F(a, b, c)... that is, F(a, b, c) = ab + bc + ca - abc

    The gradient of F is grad F(a, b, c) = (b + c - bc, a + c - ac, a + b - ab)

    The only critical point occurs where a = b = c = 2, which is outside our area of interest (namely, we want a + b + c = 2, 0 <= a <= b <= c). Thus we must only look at our function on the edges and corners... to find these, we consider the system

    1) 0 <= a <= b <= c
    2) c <= a + b
    3) a + b + c = 2

    Clearly, c <= 1, from 2) and 3). It follows that a and b are also less than 1.

    I leave the rest as an exercise </lazy>
     
  4. Oct 12, 2009 #3
    I know this is an old thread, but I recently ran into this problem and found its solution using AM-GM to be very nice. Anyone working on this who is stuck may want to expand (1-a)(1-b)(1-c) out and work from there.
     
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