A problem on maxima and rearrangements

1) By the rearrangement inequality, it is enough to prove this inequality when σ is the permutation (or, more precisely, one of the permutations) which makes the sequences (a1, a2, ..., an) and ( bσ(1), bσ(2), ..., bσ(n) ) equally sorted (because if we treat a1, a2, ..., an and b1, b2, ..., bn are constants, then this permutation σ maximizes the left hand side a1bσ(1)+a2bσ(2)+ ...+anbσ(n) of our inequality, whereas the right hand side is constant). But I don’t think this helps in solving the problem. It is actually getting the cart before the horse: The rearrangement inequality can be derived from our problem (see the remark after the solution for details). 2) The problem was conceived by me as a lemma to prove the ”combinatorialist’s Chebyshev inequality”, which states that (a1 + a2 + ...+ an) (b1 + b2 + ...+ bn) ≤ n (c1 + c2 + ...+ cn) (under the conditions of the problem) and is due to Ahlswede and Blinovsky ([1], Lecture 15, Consequences of Theorem 33, no. 2). This inequality can be derived from our problem by summing the left hand side over σ ∈ Cn (where Cn denotes the subgroup of the symmetric group Sn formed by all cyclic permutations). We leave the details to the reader. The proof in [1] is completely different.