## [數學]不等式題

### [數學]不等式題

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

skywalker

### [問題]你的問題

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

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

???

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
追求神乎其技,至高無上的數學境界!~

skywalker

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
追求神乎其技,至高無上的數學境界!~

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

http://blog.pixnet.net/ej0cl6
↑這是最近成立的數學BLOG
裡面有些幾何的東西
大家可以參觀看看呢

☆ ~ 幻 星 ~ ☆

=>(a^2+1/a^2)+(b^2+1/b^2)+(c^2+1/c^2)≧82/3

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

skywalker

tangpakchiu

skywalker

### [問題]你的問題

'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',

tangpakchiu

(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

追求神乎其技,至高無上的數學境界!~

tangpakchiu

skywalker