simple inequality

x2 + x + 1
= (x2 + x) + 1
= [x2 + x + (1/2)2] - (1/2)2 + 1
= (x + 1/2)2 + 3/4
For all real values of x, (x + 1/2)2 ≥ 0
Therefore, x2 + x + 1 = (x + 1/2)2 + 3/4 ≥ 3/4

(x2 + x + 1)(x2 - 3x + 2) ≤ 0
(x2 - 3x + 2) ≤ 0/(x2 + x + 1)
x2 - 3x + 2 ≤ 0
(x - 1)(x - 2) ≤ 0

When x < 1, (x - 1) < 0 and (x - 2) < 0, and thus (x - 1)(x - 2) > 0 (false)
When 1 ≤ x ≤ 2, (x - 1) ≥ 0 and (x - 2) ≤ 0, and thus (x - 1)(x - 2) ≤ 0 (true)
When x > 2, (x - 1) > 0 and (x - 2) > 0, and thus (x - 1)(x -2) > 0 (false)

Ans: 1 ≤ x ≤ 2
=

no solution for(x^2 + x + 1)

(x^2 - 3x + 2) ≦0
1≦x≦2

2009-01-09 22:41:50 補充：
亂碼..... -v-


no solution for(x^2 + x + 1)

(x^2 - 3x + 2)<=0
1 <= x <= 2