條件不等式

a^3 + b^3 + c^3 + 6abc
By putting c = 1 – a – b, we can obtain
P = a^3 + b^3 + (1 – a – b)^3 + 6ab(1 – a – b)
=a^3 + b^3 + (1 – 3a – 3b + 3a^2 – a^3 – 3a^2b + 3b^2 – 3ab^2 – b^3 + 6ab) + (6ab – 6a^2b – 6ab^2)
= 1 + 3a^2 + 3b^2 – 3a – 3b + 12ab – 9ab^2 – 9a^2b
= (3 – 9b)a^2 – (9b^2 – 12b + 3)a + (3b^2 – 3b + 1)
= 3(1 – 3b)a^2 – 3(3b – 1)(b – 1)a + (3b^2 – 3b + 1)
Completing square:
= 3(1 – 3b)[a^2 – (1 – b)a + (1 – b)^2/4] + 3b^2 – 3b + 1 – (1 – b)^2/4*[3(1 – 3b)]
P = 3(1 – 3b)[a – (1 – b)/2]^2 + 1/3 – (1 – 3b)^3/12 … (1)
Since a + b + c = 1, at least one of a, b, or c is <= 1/3. Without loss of generality, let’s assume b <= 1/3.
If b = 1/3, (1) becomes P = 1/3 > 1/4 … (A)
If b < 1/3 (so that 1 – 3b > 0), then by considering equation (1) as quadratic expression in a, the minimum value of P occurs when a = (1 – b)/2 with a minimum of
MIN = 1/3 – (1 – 3b)^3/12
As b < 1/3, 1 – 3b is always positive, and the minimum of MIN is when b = 0 yielding a minimum for MIN = 1/3 – 1/12 = 1/4 …(B)
Hence the absolute minimum occurs when b = 0, a = (1 – b)/2 = 1/2 and c = 1 – 0 – 1/2 = 1/2
(A) & (B) => The absolute minimum is 1/4

2009-12-14 20:05:18 補充：
實情是這樣的,上星期有人出一道題,是有關機率的,同這題的算法是一樣的,但那人在我答完之後就刪除問題.唉!
我想是Tony拔刀相助,免我白白算了!!!

2009-12-14 21:40:04 補充：
是你的問題嗎?那是匿名的啊!是說骰仔的,不規則的骰仔,不明確的數字,三骰之和是三的倍數,機率>= 1/4...

2009-12-14 23:35:56 補充：
a,b,c都受了限制 0<=a,b,c<=1
看看邊界情況:
若a=1,則b=c=0, 最大值是一
若a=0,則a^3 + b^3 + c^3 + 6abc = b^3 + c^3
= b^3 + (1 - b)^3
= 1 - 3b + 3b^2
= 3(b - 1/2)^2 + 1/4
b = 0 或 b=1 都是最大值1, b = 1/2是最小值1/4
一般的不等式都沒有這種邊界約束
若可以是負數, 當a=1, b=1, c = -1; a^3 + b^3 + c^3 + 6abc = -5

短短10分鐘....
這就是魔人布歐的恐佈實力!

短短10分鐘....

2009-12-14 19:45:23 補充：
應該係唔夠10分鐘.....

2009-12-14 21:41:09 補充：
我亦有發表過有關機率問題-.-
不過是胡亂作出來..
又是誤會一場!

2009-12-14 21:43:05 補充：
當我沒說過吧@@
我刪除有關意見..罷了-.-