MQ59 --- Inequality

(x - 2)2 (3x2 + 4x + 4) = (x - 2)2 [3(x2 + 4x/3) + 4]

= (x - 2)2 [3(x2 + 4x/3 + 4/9) + 4 - 4/3]

= (x - 2)2 [3(x2 + 2/3)2 + 8/3]

Since (x - 2)2 >= 0 and 3(x2 + 2/3)2 + 8/3 > 0, we have:

(x - 2)2 (3x2 + 4x + 4) >= 0 for all real x >= 0 with equality holds when x = 2


2012-12-17 21:14:58 補充：
W/o factorization, we can use differentiation to show it too:

f(x) = 3x⁴ - 8x³ + 16
f'(x) = 12x³ - 24x^2

f'(x) = 0 -> x = 0 or 2
When x < 0, f'(x) < 0
When 0 < x < 2, f'(x) < 0
When x > 2, f'(x) > 0

So there's a min. at x = 2 and f(2) = 0

Hence f(x) >= 0

(x⁴ + x⁴ + x⁴ + 16)/4 ≥ ∜(x⁴ ∙ x⁴ ∙ x⁴ ∙ 16) [AM ≤ GM]
3x⁴ + 16 ≥ 8³
3x⁴ - 8x³ + 16 ≥ 0

2012-12-17 21:17:14 補充：
Of course, calculus KO many of these kinds of problems...

[(x - 2)^2](3x^2 + 4x + 4)
= [(x - 2)^2][2x^2 + (x + 2)^2] ≥ 0


2012-12-15 13:55:46 補充：
Let f(x) = 3x^4 - 8x^3 + 16
f(x) = x^3(3x - 8) + 16
f(2) = 8(6 - 8) + 16 = -16 + 16 = 0
i.e. (x - 2) is a factor of f(x)

2012-12-15 14:04:03 補充：
3x^4 - 8x^3 + 16
= (x - 2)(ax^3 + bx^2 + cx + d)
= ax^4 + (b - 2a)x^3 + (c - 2b)x^2 + (d - 2c)x - 2d
i.e.
a = 3, d = -8
b - 2a = -8 -> b = -2
c - 2b = 0 -> c = -4
f(x) = (x - 2)(3x^3 - 2x^2 - 4x - 8) = (x - 2)g(x)
g(2) = 3(8) - 2(4) - 4(2) - 8 = 24 - 8 - 8 - 8 = 0
i.e. (x - 2) is a factor of g(x)

2012-12-15 14:07:35 補充：
g(x)
= 3x^3 - 2x^2 - 4x - 8
= (x - 2)(ex^2 + fx + g)
= ex^3 + (f - 2e)x^2 + (g - 2f)x - 2g
i.e.
e = 3
f - 2e = -2 -> f = -2 + 2(3) = 4
-2g = -8 -> g = 4
hence
f(x) = [(x - 2)^2](3x^2 + 4x + 4)
By the method before, f(x) ≥ 0