by rfernandez Thu May 15, 2008 7:29 pm
Very good solution by Mikus. I just want to point out that normally dividing both sides of an inequality by a variable amount is dangerous territory because of the possibility that the variable amount could be zero or a negative number. It works out well in this problem because of the constraint that 0<x<1. When we divide both sides by x^2, we know it's a positive, nonzero number. Also, when we divide both sides by (1-x), again we know it's a positive, nonzero number.
Again, nice work, Mikus!
Rey