數學證明題

1.
ab
= a(1-a)
= -a^2 + a
= - (a+1/2)^2 +1/4
<= 1/4

2.
(a+1/a)(b+1/b)
= ab + b/a + a/b + 1/ab
= ab + (a^2+b^2)/ab + 1/ab
= ab + (1-2ab)/ab + 1/ab (Because (a+b)^2 = 1, a^2+2ab+b^2 = 1, (a^2+b^2)= 1-2ab)
= ab + 2/ab -2

The function f(x) = x - 2 +2/x is decreasing for 0 < x <=1/4
Therefore, min f(x) = f(1/4) = 25/4

1. 用了中四程度 (Completing square)
2. 用了中五程度 (Differentiation)

2007-02-10 13:06:54 補充：
Himyisland8132, we need to explain why the extreme occurs at ab=1/4 and why it is a minimumGuttvennnnnnnnnn, we cannot do inequality by one-by-one substitutioni.e. x 1/x <=1/x 1 if x<=1because you cannot ensure that the extreme value of other part of function occur at x=1.

1. 設 a 及 b 為正數, 且 a+b=1.
(i) 證明 ab<=1/4
[證明ab小於等於4分之1]
a+b=1
(a+b)^2=1
a^2+b^2+2ab=1
consider
a^2+b^2
=a^2+(1-a)^2
=2a^2-2a+1
=2(a-1/2)^2+1/2
The minimum value is 1/2
So
a^2+b^2+2ab=1
1/2+2ab<=1
ab<=1/4
(ii)求(a+1/a)(b+1/b)的最小值
[求(a加 a分之1)乘(b加 b分之1)的最小值]
(a+1/a)(b+1/b)
=ab+a/b+b/a+1/ab
=ab+(a^+b^2+1)/ab
=ab+(2-2ab)/ab
=ab+2/ab-2
<=1/4+2/(1/4)-2
=25/4


2007-02-10 12:43:58 補充：
原來真係應該是大於號&gt;=1/4+2/(1/4)-2=25/4(a+1/a)(b+1/b)的最小值是25/4我原本以為出錯﹐sorry