Let a 1 , a 2 , … , a 2 0 2 0 be positive real numbers. What is the largest value of C such that
1 ≤ p < q < r ≤ 2 0 2 0 ∑ a p + a q + a r a p 2 + a q 2 + a r 2 ≥ C m = 1 ∑ 2 0 2 0 a m
holds?
Bonus: Generalize this to n instead of 2 0 2 0 and 1 ≤ i 1 < i 2 < ⋯ < i k ≤ n
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Nice, do you see the generalization to 1 ≤ i 1 < i 2 < ⋯ < i k ≤ n instead of just 1 ≤ p < q < r ≤ n ? This can be formalized in another way (which is equivalent to your approach) using QM-AM (or rather the generalized power mean inequality). The intended general result is C = ( k − 1 n − 1 ) / k
Even though it might be fairly easy to note, it would be helpful to future readers if you include why we take ( k − 1 n − 1 ) and not ( k n ) . The idea is that we're not counting k-subsets of a set with n elements but rather in how many of those subsets, a particular a i appears.
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If we replace the constraint 1 ⩽ p < q < r ⩽ n with 1 ⩽ p ⩽ q ⩽ r ⩽ n , then the answer is n 1 ( 2 n ) 2 = 4 n ( n − 1 ) 2 , with a 1 = a 2 = ⋯ = a n as usual.
Extending further, if we replace 1 ⩽ p ⩽ q ⩽ r ⩽ n with 1 ⩽ a 1 ⩽ a 2 ⩽ ⋯ ⩽ a n , then the answer is n 2 n + 2 ⋅ ( 4 n + 3 ) = 1 2 1 ( n + 1 ) 2 ( n + 2 ) ( n + 3 ) .
Is there a combinatorial interpretation for these results?
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