I guessed the max value of $c$ is $\dfrac{1389}{2}$. Is my answer right? If yes, I will post my solution.

Raja Oktovin wrote: I guessed the max value of $c$ is $\dfrac{1389}{2}$. Is my answer right? If yes, I will post my solution. No.When $a_{i}=1389-i$, then c is not bigger than $963966/2779$ which is smaller than 347. I wonder if there omit some condition,maybe the sequence is non-decreased.

Anyone notice this link?

This problem can be solved easily by compensation.

Also easily by strong induction

Change 1389 into positive integer $n $we can get answer$\frac{ n(n-1)}{4n+2}$

Stranger8 wrote: Change 1389 into positive integer $n $we can get answer$\frac{ n(n-1)}{4n+2}$ Yes and the example is easy but how did you prove that??

Very nice problem, here's a somewhat long solution for all $n$. first a lemma: if $a,b,c\geq d>0, 2c\geq a+b$ then we must have $2\sqrt{c^2-d^2}\geq \sqrt{a^2-d^2} +\sqrt{b^2-d^2}$. proof: because the inequality is homogeneous we can WLOG set $d=1$. now it is enough to prove that $$2\sqrt{(\frac{a+b}{2})^2-1}\geq \sqrt{a^2-1} +\sqrt{b^2-1}$$squaring both sides and simplifying, it remains to prove that $ab-1\geq \sqrt{(a^2-1)(b^2-1)}$ which can be seen due to Aczel's inequality (or y'know, just squaring again). keep in mind that $a,b>1$. also note that $\sum i=\frac{n(n+1)}{2}, \sum i^2=\frac{n(n+1)(2n+1)}{6}, \sum i^3=(\frac{n(n+1)}{2})^2$ Back to the problem, we will prove the problem for $c=c_n=\frac{n(n-1)}{2(2n+1)}$ (equality case is $a_i=n-i$): set $S=\sum_{i=0}^{n}ia_i^2- c_n\sum_{i=0}^{n}a_i^2$ First of all if we rearrange the $a_i$s such that $a_0\geq \dots\geq a_n$, the RHS remains the same, however the LHS decreases. so WLOG we can assume as such. now we set $a_i\rightarrow \sqrt{a_i^2-a_n^2}$. it is clear to see that $S\rightarrow S-(\frac{n(n+1)}{2}-c)a_n^2 \leq S$. also by the lemma, the new $a_i$s are still "concave". so again, we can now WLOG set $a_n=0$ Now, to make things clearer we set $x_i=a_{n-i}$. rearranging the inequality for the $x_i$s leaves us to prove that $$T=d_n \sum_{i=1}^{n} x_i^2- \sum_{i=1}^{n} ix_i^2\geq 0$$where $d_n=\frac{3n(n+1)}{2(2n+1)}$. note that the sequence $x_i$ is still concave and $2x_1\geq x_2$ (beacuse $a_n=0$). now we prove that $\frac{x_i}{i}\geq \frac{x_{i+1}}{i+1}$. let $k$ be the smallest number that it isn't so (keep in mind that for $i=1$ it is correct), then we must have: $$2kx_k\geq k(x_{k+1}+x_{k-1}) \geq (k+1)x_k +kx_{k-1} \Rightarrow \frac{x_{k-1}}{k-1}\geq \frac{x_{k}}{k}$$a contradiction. call such a sequence, "nice". we will now prove that $T\geq 0$ for all "nice" sequences $x-i$ by induction on $n$, the base case is clear. let $k$ be the smallest number that $x_k<kx_1$ (we know that $x_i\leq ix_1$). define: $$A=d_n \sum_{i=1}^{k-1} x_i^2- \sum_{i=1}^{k-1} ix_i^2, B=d_n \sum_{i=k}^{n} x_i^2- \sum_{i=k}^{n} ix_i^2$$realizing that $d_n>d_{n-1}$ it is clear that $A\geq0$. if $B\geq0$ then $T=A+B\geq 0$ and we are done, so let $B<0$. then setting $x_k,\dots x_n\rightarrow ax_k,\dots ax_n$ where $a=\frac{k}{x_k}>1$ we can see that $T'=A+a^2B<A+B=T$. the sequence being nice, one can easily conclude the new sequence defined as $x_1,\dots,x_{k-1},ax_k,\dots ax_n$ is nice as well. (keep in mind that $\forall i<k$ we have $x_i=ix_1$). Now start from any nice sequence and perform the aforementioned operation. note that each move our $T$ decreases. Either it stops when at some point the $B$ we create is nonnegative thus the $T$ at that point is also positive, proving our first $T$ was also nonnegative. or we reach the point that for all $i$ we have $a_i=ia_1$. but at this point using the sum identities introduced at the start we get $T=0$ so the first $T$ was also nonnegative. Either way we are done.

An easy way to prove.

Let $n=1389.$ First, let $a_k=n-k,$ then $\{a_n\}$ is a concave sequence, then $$c\le\dfrac{\sum\limits_{i=0}^ni(n-i)^2}{\sum\limits_{i=0}^n(n-i)^2}=\dfrac{n\sum\limits_{i=0}^ni^2-\sum\limits_{i=0}^ni^3}{n(n+1)(2n+1)/6}=\frac{n^2(n+1)(2n+1)/6-n^2(n+1)/4}{n(n+1)(2n+1)/6}=\frac{n(n-1)}{2(2n+1)}.$$Define $c_0=\frac{n(n-1)}{2(2n+1)},$ Now we only need to prove: ${\sum\limits_{i=0}^nia_i^2\ge c_0\sum\limits_{i=0}^na_i^2}$ holds for any concave sequence $\{a_n\}.$ Define $k=\lfloor c_0\rfloor,$ $\{b_t\}_{t=0}^n:$ $b_t=\dfrac{a_k}{n-k}\cdot (n-t).$ From $a_k=n_k,a_n\ge 0=b_n,$ using the concavity we have $b_t\ge a_t,\forall t\le k;$ $b_t\le a_t,\forall t\ge k.$ Therefore $$\sum_{i=k+1}^n(i-c_0)a_i^2\ge \sum_{i=k+1}^n(i-c_0)b_i^2,\quad\sum_{i=0}^k(c_0-i)b_i^2\ge\sum_{i=0}^k(c_0-i)a_i^2.$$Also $\sum\limits_{i=k+1}^n(i-c_0)b_i^2=\sum\limits_{i=0}^k(c_0-i)b_i^2.$ Then $$\sum_{i=k+1}^n(i-c_0)a_i^2\ge\sum_{i=0}^k(c_0-i)a_i^2\Rightarrow{\sum\limits_{i=0}^nia_i^2\ge c_0\sum\limits_{i=0}^na_i^2},$$and we are done.$\hfill\Box$