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發表 skywalker 於 星期日 一月 21, 2007 6:49 pm

#ed_op#DIV#ed_cl#其實我應該寫的更清楚的#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#等號成立於a^2+1/a^2=b^2+1/b^2=c^2+1/c^2,a=b=c=1/3時#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為a+b+c=1,所以我後面才會用1/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#否則只要a=b=c,a^2+1/a^2=b^2+1/b^2=c^2+1/c^2就會成立#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#我好像寫的太簡略了#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

[問題]我的問題

發表 tangpakchiu 於 星期日 一月 21, 2007 6:26 pm

#ed_op#DIV#ed_cl#對...近來常常看錯東西....考試also...thank you~~~~#ed_op#/DIV#ed_cl#

發表 宇智波鼬 於 星期日 一月 21, 2007 6:20 pm

看清楚喔...
(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)≧82/3已經是整理過的了.
原式應為
(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)+2+2+2≧100/3
所以移項後證明(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)≧82/3即可.

但是後部份的說明是你必須已確定a=b=c時為最小值...否則不能這樣用.
另外...有沒有發現你從頭到尾都未使用到a+b+c=1?
既然題目有給,就代表一定需要用到.

[問題]你的問題

發表 tangpakchiu 於 星期日 一月 21, 2007 5:44 pm

#ed_op#DIV#ed_cl#'所以(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)≧3*(3^2+1/3^2)=82/3',#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#雖然能證明式子大過等於82/3...但你能確保它也大過83/3嗎???#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#或者可以由不等式數線上看,線上有82/3,100/3,0這3點(0在線的最左邊)。題目是要證明由100/3到0的範圍都成立.your answer only proves that 由82/3到0到正確,但由100/3到82/3這段呢??還沒有證明啊....#ed_op#/DIV#ed_cl#

發表 skywalker 於 星期日 一月 21, 2007 2:45 pm

#ed_op#DIV#ed_cl#不好意思,我不大懂你的意思,可以明確的指出來嗎#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#我自己本身也不太確定我這樣可不可以#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為感覺怪怪的#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#不過不知道錯在哪裡#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

[問題]你的問題

發表 tangpakchiu 於 星期日 一月 21, 2007 2:12 pm

#ed_op#DIV#ed_cl#應該是有問題的...因為你的證明不夠strong....如你雖然能證明x>2,但不代表x>3,因x可能是2.5.#ed_op#/DIV#ed_cl#

發表 skywalker 於 星期六 一月 20, 2007 11:26 pm

#ed_op#DIV#ed_cl#這是我的作法,不知道對不對#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為(a+1/a)^2+(b+1/b)^2+(c+1/c)^2≧100/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=>(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)≧82/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#只要證明上式成立就可以了#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由AM-GM#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#≧3*三次根號√(a^2+1/a^2)(b^2+1/b^2)(c^2+1/c^2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#等號成立於a^2+1/a^2=b^2+1/b^2=c^2+1/c^2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此時由a+b+c=1,a=b=c=1/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)≧3*(3^2+1/3^2)=82/3 ##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

發表 ☆ ~ 幻 星 ~ ☆ 於 星期六 一月 20, 2007 8:19 pm

a+b+c為定值時
當a=b=c,abc會有最大值
此時是1/27


由柯西不等式
(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]≧(a+b+c+1/a+1/b+1/c)^2

3[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]≧(1+1/a+1/b+1/c)^2

3[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]≧(1+3*三次根號(1/abc))^2

3[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]≧(1+9)^2

(a+1/a)^2+(b+1/b)^2+(c+1/c)^2≧100/3

發表 宇智波鼬 於 星期六 一月 20, 2007 8:03 pm

原來打錯題目阿...= =

設a+1/a=x b+1/b=y c+1/c=z
則3(x^2+y^2+z^2)>=x^2+y^2+z^2+2xy+2yz+2xz=(x+y+z)^2
所以x^2+y^2+z^2>=(x+y+z)^2/3
而a+b+c/3=1/3>=3/(1/a+1/b+1/c)
所以1/a+1/b+1/c>=9
a+1/a+b+1/b+c+1/c>=1+9=10
所以(a+1/a)^2+(b+1/b)^2+(c+1/c)^2=x^2+y^2+z^2
>=(x+y+z)^2/3=(a+1/a+b+1/b+c+1/c)^2/3=100/3

發表 skywalker 於 星期六 一月 20, 2007 7:06 pm

#ed_op#DIV#ed_cl#抱歉我打錯了.....#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#是a+b+c=1,這樣才對0.0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#

發表 宇智波鼬 於 星期六 一月 20, 2007 3:56 pm

This inequality is weird. Are you sure it's the original one?
If you assume that a=1 b=1 c=1, then,(a+1/a)^2+(b+1/b)^2+(c+1/c)^2=12<100/3.

My method:
(a+1/a)^2+(b+1/b)^2+(c+1/c)^2
>=3[(a+bc)(b+ac)(c+ab)]^(2/3)
>=3(a^2+b^2+c^2+a^2b^2+b^2c^2+a^2c^2+abc+abc)^(2/3)
>=3[8(abc)^(1/8)]^(2/3)=3(8)^(2/3)=12

[問題]你的問題

發表 ??? 於 星期六 一月 20, 2007 2:32 pm

#ed_op#DIV#ed_cl#a,b,c是正數且abc=1,試證(a+1/a)^2+(b+1/b)^2+(c+1/c)^2≧100/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#設(a+1/a)^2+(b+1/b)^2+(c+1/c)^2=x#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x≧(a+1/a+b+1/b+c+1/c)^2/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x≧(3(abc)^(1/3)+1/3(abc)^(1/3))^2/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x≧100/3#ed_op#/DIV#ed_cl#

[數學]不等式題

發表 skywalker 於 星期六 一月 20, 2007 12:08 am

#ed_op#DIV#ed_cl#a,b,c是正數且abc=1,試證(a+1/a)^2+(b+1/b)^2+(c+1/c)^2≧100/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl#