$\dfrac{ x_{1} }{ x_{2}+2 x_{3}+...+(n-1) x_{n} } + \dfrac{ x_{2} }{ x_{3}+2 x_{4}+...+(n-1) x_{1} } +...+ \dfrac{ x_{n} }{x_{1}+2x_{n2}+...+(n-1) x_{n-1} }\geq$ $\dfrac{2}{n-1} $
@vietfrog: Gõ latex bạn nha!
Edited by vietfrog, 13-09-2011 - 00:17.
Bất đẳng thức
Started By lovely_kunju_1803, 13-09-2011 - 00:02
#1
Posted 13-09-2011 - 00:02
lovely_kunju_1803
-
- Thành viên
-
- 31 posts
Binh nhất
cho $x_{1 }, x_{2},..., x_{n}$ >0,Chứng minh rằng:
$\dfrac{ x_{1} }{ x_{2}+2 x_{3}+...+(n-1) x_{n} } + \dfrac{ x_{2} }{ x_{3}+2 x_{4}+...+(n-1) x_{1} } +...+ \dfrac{ x_{n} }{x_{1}+2x_{n2}+...+(n-1) x_{n-1} }\geq$ $\dfrac{2}{n-1} $
@vietfrog: Gõ latex bạn nha!
$\dfrac{ x_{1} }{ x_{2}+2 x_{3}+...+(n-1) x_{n} } + \dfrac{ x_{2} }{ x_{3}+2 x_{4}+...+(n-1) x_{1} } +...+ \dfrac{ x_{n} }{x_{1}+2x_{n2}+...+(n-1) x_{n-1} }\geq$ $\dfrac{2}{n-1} $
@vietfrog: Gõ latex bạn nha!
....The key to success is to focus our conscious mind on things we desire not things we fear....
...................................................
.......................
No name. It 's me
#2
Posted 13-09-2011 - 00:30
alex_hoang
-
- Hiệp sỹ
-
- 1152 posts
Thượng úy
Nó không khó đâu bạn hoàn toàn có thể cm bởi Cauchy Schwarzcho $x_{1 }, x_{2},..., x_{n}$ >0,Chứng minh rằng:
$\dfrac{ x_{1} }{ x_{2}+2 x_{3}+...+(n-1) x_{n} } + \dfrac{ x_{2} }{ x_{3}+2 x_{4}+...+(n-1) x_{1} } +...+ \dfrac{ x_{n} }{x_{1}+2x_{n2}+...+(n-1) x_{n-1} }\geq$ $\dfrac{2}{n-1} $
@vietfrog: Gõ latex bạn nha!
Edited by alex_hoang, 13-09-2011 - 00:30.
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
