Expert: Sherman D. - 10/28/2007

Question
If X is real and P=(3(X2+1))/(2X-1), prove that P2-3(P+3)>=0.

Answer
if by this you mean

P = (3(x^2 + 1))/(2x - 1)
P^2 - 3(P + 3) >= 0
P^2 - 3P - 9 >= 0

(3(x^2 + 1)/(2x - 1))^2 = 9((x^2 + 1)/(2x - 1))^2

3(3(x^2 + 1)/(2x - 1)) = 9((x^2 + 1)/(2x - 1))

9((x^2 + 1)/(2x - 1))^2 - 9((x^2 + 1)/(2x - 1)) - 9 >= 0

((x^2 + 1)/(2x - 1))^2 - ((x^2 + 1)/(2x - 1)) - 1 >= 0

((x^2 + 1)^2 - ((x^2 + 1)(2x - 1)) - (2x - 1)^2)/((2x - 1)^2) >= 0

(x^4 + 2x^2 + 1 - (2x^3 - x^2 + 2x - 1) - (4x^2 - 4x + 1))/((2x - 1)^2) >= 0

(x^4 + 2x^2 + 1 - 2x^3 + x^2 - 2x + 1 - 4x^2 + 4x - 1)/((2x - 1)^2) >= 0

(x^4 - 2x^3 - x^2 + 2x + 1)/((2x - 1)^2) >= 0

using www.quickmath.com

x < (1/2)
or
x > (1/2)

so

P^2 - 3(P + 3) >= 0 is only true if x < (1/2) or x > (1/2)