Hint for Dificult Problem

Last week I posted a difficult problem and only had one person even try. The problem was: Assume a, b, and c are positive real numbers. Show:

I hinted in a comment that there is a way of limiting the problem without changing the problem. The way to do this is by exploiting the symmetry of the problem. Note that we can freely interchange the variables and the problem remains exactly the same, thus we can assume a is the largest, b the middle value, and c the smallest, that is we may assume:
a >= b >= c > 0.

Second, note that we can multiply all 3 numbers by any positive number t > 0. When we do this:

It's easy to see the t's will all cancel and we're left with our original problem. Thus we can let a' = (a* 1/c), b' = (b* 1/c), and c' = (c*1/c) = 1, and this will not change the problem in anything more than appearance. But it will simplify to:

with the assumption a' >= b' >= 1.

This makes the problem a bit easier by eliminating one of the independent variables and also eliminates having to worry about cases where one of the numbers may be very close to zero.

I had hoped someone would come along with a clever solution for this problem. At first, I had thought it could be solved by algebra alone, but the solution I came up with uses algebra to eliminate most choices for a & b, and then needs a bit of calculus to show the statement is true for a small domain. Because the minimum value of 3/2 is actualized when all three values are equal, I am pretty sure you'll need calculus at some point. No matter how close you choose to estimate things, an estimate will never get you to exactly 3/2, and precision is the advantage calculus offers over other simpler techniques of mathematics.

Good luck on the second try!