Putnam and beyond prob 124

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>

 

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.