x^2-6x+11 = x^2 - 6x + 9 + 2 = (x - 3)^2 + 2
Therefore a = -3 and b = 2, and its mimimum is 2 when x = 3
For 1/(x^2-6x+11) = 1/[(x-3)^2+2], the maximum is 1/2 when x = 3 and is positive since (x-3)^2+2 >=2 > 0.
In conclusion, 0 < 1/(x^2-6x+11) <= 1/2.