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The Chebyshev Sum Inequality

If a1 ≥ a2 ≥ ... ≥ an and b1 ≥ b2 ≥ ... ≥ bn then

n  n

akbk (  n

ak )(  n

bk )

The proof is that the RHS can be written as the sum of n different sums,
   (a1b1 + a2b2 + ... + an-1bn-1 + anbn) + 
   (a1b2 + a2b3 + ... + an-1bn + anb1) + 
   (a1bn + a2b1 + ... + an-1bn-2 + anbn-1)
The first of these n sums is at least as big as each of the others by the rearrangement inequality, and n times this first sum is the LHS, so the LHS is at least as big as the RHS.


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