Show that \((x,a_k \in [1,1], \sum_{k=1}^n a_k = 0) \Rightarrow (\sum_{k=1}^n xa_k \leq n)\)

03122010, 09:42 AM
(This post was last modified: 03182010 06:45 PM by elim.)
Post: #1




Show that \((x,a_k \in [1,1], \sum_{k=1}^n a_k = 0) \Rightarrow (\sum_{k=1}^n xa_k \leq n)\)
And what's the background or interpretation of the inequality?


03182010, 06:59 PM
(This post was last modified: 03182010 09:35 PM by elim.)
Post: #2




RE: Show that \((x,a_k \in [1,1], \sum_{k=1}^n a_k = 0) \Rightarrow (\sum_{k=1}^n xa_k \leq n)\)
Let \(x \in [1,1]\) be the one that makes \(\sum_{k=1}^n xa_k\) attain its maximum.
Let \(\left\{b_i\right \}, \left\{c_j\right \}\) be the subsets of \(\left\{a_k\right \}\) such that \(xa_k > 0 (\leq 0)\) respectively. Write \(p\) and \(q\) as the count of \(\left\{b_i\right \}, \left\{c_j\right \}\) respectively and we then have \(\sum_{k=1}^n xa_k = \sum_1^p (xb_i) + \sum_1^q (c_jx) =x(pq)\sum_1^p b_i+\sum_1^q c_j \) If \(p>q\) then \(x = 1\), \(\sum_{k=1}^n xa_k = pq +2 \sum_1^q c_j \leq pq +2q=n\) Otherwise \(p \leq q\), \(x = 1\) and \(\sum_{k=1}^n xa_k = qp 2 \sum_1^p b_i \leq qp +2p=n\) 

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