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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
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
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?
03-18-2010, 06:59 PM
Post: #2
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
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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