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Show that \((x,a_k \in [-1,1], \sum_{k=1}^n a_k = 0) \Rightarrow (\sum_{k=1}^n |x-a_k| \leq n)\)
03-12-2010, 09:42 AM
Post: #1
Show that \((x,a_k \in [-1,1], \sum_{k=1}^n a_k = 0) \Rightarrow (\sum_{k=1}^n |x-a_k| \leq n)\)
And what's the background or interpretation of the inequality?
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03-18-2010, 06:59 PM
Post: #2
RE: Show that \((x,a_k \in [-1,1], \sum_{k=1}^n a_k = 0) \Rightarrow (\sum_{k=1}^n |x-a_k| \leq n)\)
Let \(x \in [-1,1]\) be the one that makes \(\sum_{k=1}^n |x-a_k|\) attain its maximum.
Let \(\left\{b_i\right \}, \left\{c_j\right \}\) be the subsets of \(\left\{a_k\right \}\) such that \(x-a_k > 0 (\leq 0)\) respectively.
Write \(p\) and \(q\) as the count of \(\left\{b_i\right \}, \left\{c_j\right \}\) respectively and we then have \(\sum_{k=1}^n |x-a_k| = \sum_1^p (x-b_i) + \sum_1^q (c_j-x) =x(p-q)-\sum_1^p b_i+\sum_1^q c_j \)
If \(p>q\) then \(x = 1\), \(\sum_{k=1}^n |x-a_k| = p-q +2 \sum_1^q c_j \leq p-q +2q=n\)
Otherwise \(p \leq q\), \(x = -1\) and \(\sum_{k=1}^n |x-a_k| = q-p -2 \sum_1^p b_i \leq q-p +2p=n\)
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