數學測驗勁難題.......help下me...........

consider
(a-b/2)^2 = a^2-ab+b^2/4 -----(1)
3(b/2-3)^2 = 3b^2/4-9b+27 ----(2)
(c-4)^2 =c^2 -8c +16 -----(3)
(1)+(2)+(3), we have
(a-b/2)^2+3(b/2-3)^2+(c-4)^2=a^2+b^2+c^2+43 -ab -9b -8c
ab+9b+8c = a^2+b^2+c^2 +43 -(red part＞＝0)
ab+9b+8c＞a^2+b^2+c^2+43
are you sure the integer is 42 not 43, please reply me
(to be continued)




2007-11-07 19:18:48 補充：
since 43 42, thenab 9b 8c＞a^2 b^2 c^2 43 a^2 b^2 c^2 42

2007-11-07 19:20:23 補充：
since 43＞42, thenab＋9b＋8c＞a^2＋b^2＋c^2＋43＞a^2＋b^2＋c^2＋42

2007-11-07 19:21:21 補充：
hope I can help you!

2007-11-07 22:16:21 補充：
sorry I just made a mistake in hurryab＋9b＋8c＝ a^2＋b^2＋c^2＋43－(red part＞＝0)thereforeab+9b+8c＜＝a^2＋b^2＋c^2＋43

2007-11-07 22:16:34 補充：
your inequality is not true, you can just by putting a,b andc equal to 1, then you will find you inequality cannot hold. Therefore, I believe you make mistakes(also including writing 42 instead of 43).

2007-11-12 00:09:05 補充：
因為ab+9b+8c = a^2+b^2+c^2 +43 -(a-b/2)^2+3(b/2-3)^2+(c-4)^2所以，若要ab+9b+8c＞a^2+b^2+42的話那麼red part就必須等於０所以c-4=0, b/2-3=0, a-b/2=0所以 c=4, b=6, a=3

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

3(b/2-3)^2 = 3b^2/4-9b+27 ----(2)

(c-4)^2 =c^2 -8c +16 -----(3)

(1)+(2)+(3), we have

(a-b/2)^2+3(b/2-3)^2+(c-4)^2=a^2+b^2+c^2+43 -ab -9b -8c

ab+9b+8c = a^2+b^2+c^2 +43 -(red part＞＝0)

ab+9b+8c＞a^2+b^2+c^2+43

are you sure the integer is 42 not 43, please reply me