Originally Posted by luckyvan
Since a,b,c are all positive then
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.
I have tried to solve it a bit different.
Putting abc instead of 1 (because 1=abc) we get
Now, lets consider next cases:
If a,b,c are all equal to 1 then is true.
If one of them is equal to 1 and others not (for example let be )
we would get:
Knowing that then is equal to:
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
is true.
Hope this solve the problem.
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