Given x k , y k , z k > 0 for all k satisfy the three summations:
k = 1 ∑ 2 0 1 5 x k k = 1 ∑ 2 0 1 5 y k k = 1 ∑ 2 0 1 5 z k = = = 5 0 1 6 9 9 6 1
What is the minimum value of k = 1 ∑ 2 0 1 5 ( x k y k z k ) ?
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I thought Chebyshev's Inequality works even for more than 2 sequences. I used that and got the minimum 2 . Guess I was wrong.
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Yeah, I was surprised too that it doesn't generalize.
Same mistake, dude!!
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Let x 1 = 5 0 − α and x k = 2 0 1 4 α for 2 ≤ k ≤ 2 0 1 5 for some α > 0 .
Let y 2 = 1 6 9 − α and y 1 = y k = 2 0 1 4 α for 3 ≤ k ≤ 2 0 1 5 .
Let z 3 = 9 6 1 − α and z 1 = z 2 = z k = 2 0 1 4 α for 4 ≤ k ≤ 2 0 1 5 .
Then the initial three sum conditions are satisfied, and we find that
S = k = 1 ∑ 2 0 1 5 x k y k z k = ( 5 0 + 1 6 9 + 9 6 1 − 3 α ) ( 2 0 1 4 α ) 2 + 2 0 1 2 ( 2 0 1 4 α ) 3 .
We can then make S arbitrarily small by making α arbitrarily small, but since α can never actually be 0 , (since x k , y k , z k > 0 for all k ), S can never be zero. Thus there is no minimum.
Note that, since all the terms of S exceed 0 by the given conditions, there is no way for S to equal 0 . But by the above argument, if S did have a minimum associated with some α value, then we could just choose a smaller α value in order to find a "smaller minimum", hence giving rise to a contradiction. Thus we can conclude that there is no minimum. We could get more formal here, but this is the general concept at work.