### + / -檢視主題

#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#&nbsp;#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#&nbsp;#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#&nbsp;#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#&nbsp;#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

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

(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.

(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)+2+2+2≧100/3

### [問題]你的問題

#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#&nbsp;#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#&nbsp;#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#&nbsp;#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

#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&gt;2,但不代表x&gt;3,因x可能是2.5.#ed_op#/DIV#ed_cl#
#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#&nbsp;#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#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=&gt;(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#&nbsp;#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#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由AM-GM#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#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#&nbsp;#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#&nbsp;#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#&nbsp;#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#
a+b+c為定值時

(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

a+1/a+b+1/b+c+1/c>=1+9=10

>=(x+y+z)^2/3=(a+1/a+b+1/b+c+1/c)^2/3=100/3
#ed_op#DIV#ed_cl#抱歉我打錯了.....#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#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#&nbsp;#ed_op#/DIV#ed_cl#
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

#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

#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#