Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help on simple-looking ineqality

  1. Apr 6, 2006 #1
    Dear all,

    I come across to a simple-looking ineqality. But I cann't prove it for quite a long time. Could any body give a hint? Thanks a lot!

    [tex]
    2[(n-1) \sum_{j=1}^n r_j^2 -(n-2) r_n^2] \geq (\sum_{j=1}^n r_j)^2
    [/tex]

    where [tex]n\geq 2, \forall r_j \geq 0, j=1,2,\cdots,n[/tex].
     
  2. jcsd
  3. Apr 6, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you saying that you do not understand why (n-3)2rn2 should be non-negative?
     
    Last edited: Apr 7, 2006
  4. Apr 7, 2006 #3
    Thanks for your sugestions. However, I still don't know how did you get to the simpification of [tex](n-3)^2r_n^2[/tex].

    According to your sugestions, I got the following proof:

    [tex]
    2[(n-1)\sum_{j=1}^n r_j^2 -(n-2) r_n^2]
    =2[(n-1)\sum_{j=1}^{n-1} r_j^2 + r_n^2]
    \geq 2[(\sum_{j=1}^{n-1} r_j)^2 + r_n^2]
    \geq (\sum_{j=1}^{n-1} r_j)^2 + r_n^2 + 2r_n \sum_{j=1}^{n-1} r_j
    =(\sum_{j=1}^{n} r_j)^2
    [/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help on simple-looking ineqality
Loading...