條件不等式(續篇)

(a+b+c)^3=1

(a+b)^3+ 3c (a+b)^2+ 3c ^2(a+b)+c^3=1

a^3+3ba^2+3ab^2+b^3+3ca^2+6abc+3cb^2+ 3ac ^2+3bc^2+c^3=1

a^3+b^3+c^3=1-6abc-3(ba^2+ab^2+ca^2+cb^2+ac^2+bc^2)

a^3+b^3+c^3+6abc=1-3(ba^2+ab^2+ca^2+cb^2+ac^2+bc^2)


a^3+b^3+c^3+6abc=1-3(ba^2+ab^2+ca^2+cb^2+ac^2+bc^2)<=1-18abc
(因為ab^2+bc^2+ca^2>=3abc,ba^2+cb^2+ac^2>=3abc)By AM>=GM


a^3+b^3+c^3+6abc<=1-18abc

由命題可見,a^3+b^3+c^3+6abc<=1(比上述大)

a,b,c最少其中一個係零,最多當然係2個零,才等號成立


有3個情況
1)當其中一個未知數係零,另外2個不相等時
x^3+y^3=1
即x,y可以係任一實數,且x<1,y<1

2)當其中一個未知數係零,另外2個相等時
即2k^3=1
k=1/2^(1/3)

3)當其中二個未知數係零,
m^3=1
m=1










2009-12-18 12:11:11 補充：
我呢個做法係咪簡易好多呢

P = 3(1 – 3b)[a – (1 – b)/2]^2 + 1/3 – (1 – 3b)^3/12
Without loss of generality, we can assume b>1/3 so that (3b - 1)>0
P = -3(3b – 1)[a – (1 – b)/2]^2 + 1/3 + (3b – 1)^3/12

2009-12-30 20:58:10 補充：
By considering variation in a, maximum value is when a = (1-b)/2 and the max value is
1/3 + (3b – 1)^3/12
The absolute max is when b is max (=1) so max = 1/3 + (3 - 1)^3/12 = 1
This happens when b = 1, a = (1 - 1)/2 = 0 and c = 1 - 0 - 1 = 0

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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 … (A)

If b < 1/3 (so that 1 – 3b > 0), then by considering equation (1) as quadratic expression in a, ∂P/∂a=6(1-3b)[a-(1-b)/2]. If a < (1-b)/2, then ∂P/∂a < 0 If a > (1-b)/2, then ∂P/∂a > 0. So the maximum value of P is P(0) or P(1).

For a=1 , P(1)=1. For a=0, P(0)=b^3+c^3=(b+c)^3-3bc(b+c)=1-3bc(b+c)<1. So P < 1 ... (B)

Combine (A) and (B), we have a^3 + b^3 + c^3 + 6abc <= 1