How prove $\sum_{1\le i

Question

Question:

let $$0\le a_{1}\le a_{2}\le\cdots\le a_{n}\le 1$$ show that $$\sum_{1\le i<j\le n}(a_{j}-a_{i}+1)^2+4\sum_{i=1}^{n}a^2_{i}\le \begin{cases} \dfrac{5n^2+6n+4}{4}&n=2k\\ \dfrac{5n^2+6n+5}{4}&n=2k+1 \end{cases}$$

This problem is from Maths exam test simulation.and I fell this sum can't deal it.Thank you

trivially, this is bounded by $2n^2$. You are trying to improve to slightly more than $\tfrac{5}{4}n^2$ using correlations in the $a_1, \dots, a_n$. — cactus314, May 09 '14 at 12:43
If I set $a_1 = \dots a_n = 0$ the left side is $\tfrac{1}{2}n^2$. — cactus314, May 09 '14 at 12:46

Calvin Lin · Answer 1 · 2014-05-11T13:30:16.127

2

Hint: Fix k. Assume that for $i\neq k$, $a_i$ is fixed. We want to maximize the LHS. It is an upwards facing quadratic in $a_k$, so attains it's maximum at the endpoints of the interval. Thus $a_k= a_{k-1},a_{k+1}$.

Hence, all of the $a_i$ are either 1 or 0, so you now have n+1 equations to evaluate to find the maximum. The max is achieved when they are split 'somewhat evenly', which explains why the even and odd case have different bounds.

Not at computer, can provide more details later if needed. But you should have enough to work it out.

edited May 11 '14 at 13:30

answered May 09 '14 at 13:15

Calvin Lin

68,864

if you post you solution,maybe have some +1 – math110 May 09 '14 at 14:44
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@math110 That's alright, I don't need the +1. If you want more details, you can work on this similar problem. – Calvin Lin May 10 '14 at 13:09
Hello, you link only have problem,not solution? https://brilliant.org/community-problem/if-only-i-could-see-in-81-dimensions-this-would-be/?group=7e8JdfrqPKKg – math110 May 11 '14 at 13:57
@chinamath I wrote up a solution. – Calvin Lin May 11 '14 at 13:57
@chinamath You will need to log in to view the solution discussion. – Calvin Lin May 11 '14 at 14:16

How prove $\sum_{1\le i

1 Answers1