Power mean with counting

Algebra Level 5

Let a 1 , a 2 , , a 2020 a_1,a_2,\ldots,a_{2020} be positive real numbers. What is the largest value of C C such that

1 p < q < r 2020 a p 2 + a q 2 + a r 2 a p + a q + a r C m = 1 2020 a m \sum_{1\leq p\lt q\lt r\leq 2020}\frac{a_p^2+a_q^2+a_r^2}{a_p+a_q+a_r}\geq C\sum_{m=1}^{2020}a_m

holds?


Bonus: Generalize this to n n instead of 2020 2020 and 1 i 1 < i 2 < < i k n 1\leq i_1\lt i_2\lt\dots\lt i_k\leq n


The answer is 679057.

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1 solution

Ritabrata Roy
Nov 14, 2020

Nice, do you see the generalization to 1 i 1 < i 2 < < i k n 1\leq i_1\lt i_2\lt\dots\lt i_k\leq n instead of just 1 p < q < r n 1\leq p\lt q\lt r\leq 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 = ( n 1 k 1 ) / k C=\binom{n-1}{k-1}/k

Even though it might be fairly easy to note, it would be helpful to future readers if you include why we take ( n 1 k 1 ) \binom{n-1}{k-1} and not ( n k ) \binom nk . 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 a_i appears.

Prasun Biswas - 6 months, 4 weeks ago

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If we replace the constraint 1 p < q < r n 1 \leqslant p < q < r \leqslant n with 1 p q r n 1 \leqslant p \leqslant q \leqslant r \leqslant n , then the answer is 1 n ( n 2 ) 2 = n ( n 1 ) 2 4 , \frac1n \binom{n}2 ^2 = \frac{n (n-1)^2}4 , with a 1 = a 2 = = a n a_1 = a_2 = \cdots = a_n as usual.

Extending further, if we replace 1 p q r n 1 \leqslant p \leqslant q \leqslant r \leqslant n with 1 a 1 a 2 a n 1 \leqslant a_1 \leqslant a_2 \leqslant \cdots \leqslant a_n , then the answer is 2 n + 2 n ( n + 3 4 ) = 1 12 ( n + 1 ) 2 ( n + 2 ) ( n + 3 ) . \frac{2n+2}n \cdot \binom{n+3}4 = \frac 1{12} (n+1)^2 (n+2)(n+3) .

Is there a combinatorial interpretation for these results?

Pi Han Goh - 6 months, 3 weeks ago

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