if I am not wrong, we can proceed by induction. For two variable case it's trivial. Suppose it's true for k variables. We define an operation on $a_{k+1}$ as a replacement of $a_1,...,a_k$ by $a_1 + \frac{a_{k+1}}{k}$ and so forth. Since $\sum a_1 + \frac{a_{k+1}}{k} = 0$, by induction hypothesis, $\frac{k}{3} ((a_1 - a_2)^2 + ... + (a_{k-1} - a_k)^2 \ge \max \left(a_1 + \frac{a_{k+1}}{k}\right)^2$ Similarly we have the analogous inequalities by operations on $a_1,...,a_k$, sum them up and if we let $S = (a_1 - a_2)^2 + ... + (a_k - a_{k+1})^2$, we consider a variable $a_1$ (and similarly, for $a_2,...,a_{k+1}$, $\frac{k(k-1)}{3} S \ge \left(a_1 + \frac{a_2}{k}\right)^2 + \left(a_1 + \frac{a_3}{k}\right)^2 + ... + \left(a_1 + \frac{a_{k+1}}{k}\right)^2$ By Cauchy-Schwarz, $\ge \frac{1}{k} \left(k a_1 + \frac{a_2 + ... + a_{k+1}}{k} \right)^2 = \frac{(k^2-1)^2}{k^3}a_1^2$ Thus we have $S \ge \frac{3(k-1)}{k^4}(k+1)^2a_1^2$ Now, $\frac{k+1}{3}S \ge \frac{(k+1)^3(k-1)}{k^4} a_1^2$, so it remains to show $(k+1)^3(k-1) \ge k^4$, which is equivalent to $2k^3 \ge 2k + 1$, which is true for $k \ge 2$. As $\frac{k+1}{3} S \ge a_i^2$ for all $i = 1,...,{k+1}$, the induction is completed. QED

the original problem is: $\max_{1 \le i \le k} (a_i^2)\leq [(a_1-a_2)^2+(a_2-a_3)^2+...+(a_{k-1}-a_k)^2]\frac{k}{3}$ same condition.

to soarer :you surely gived a wonderful solution. Have you ever had a chance to view the official answer which is wonderful as well. I consider the official answer as a little weird to me for it's quite beyond me. Would you try to have a glance at it and give your comments? I'd also like to see how you managed to tackle the problem via your approach,I mean,how you get the idea of induction and so forth?

What's the other solution?

Do you have a chance to view the High School Mathematics (a Chinese magazine)? If so you you will find the very answer.

mysteryfigure wrote: Do you have a chance to view the High School Mathematics (a Chinese magazine)? If so you you will find the very answer. No...

Soarer wrote: $\frac{k(k-1)}{3}S \ge \left(a_{1}+\frac{a_{2}}{k}\right)^{2}+\left(a_{1}+\frac{a_{3}}{k}\right)^{2}+...+\left(a_{1}+\frac{a_{k+1}}{k}\right)^{2}$ I think this is mistaken. There's no way to get that sum.

Summerburn wrote: Soarer wrote: $\frac{k(k-1)}{3}S \ge \left(a_{1}+\frac{a_{2}}{k}\right)^{2}+\left(a_{1}+\frac{a_{3}}{k}\right)^{2}+...+\left(a_{1}+\frac{a_{k+1}}{k}\right)^{2}$ I think this is mistaken. There's no way to get that sum. ?what do you mean?

How that line comes? I can't see the left side being $\frac{k(k-1)}{3}S$

but it's obvious...

I don't understand this part, too.

As Soarer did,we will prove that $ a_i^2\leq\frac{n}{3}((a_1-a_2)^2+(a_2-a_3)^2+\dots+(a_{n-1}-a_n)^2)$,for all $ 1\leq i\leq n$. Define $ d_i=a_i-a_{i+1}$,for all $ 1\leq i\leq n$ and $ a_{n+1}=a_1$.Obviously $ d_1+d_2+\dots+d_n=0$. Also it is easy to understand that $ d_1+2d_2+3d_3+\dots+nd_n=a_1+a_2+\dots+a_n-na_{1}$,hence $ a_1=-\frac{d_1+2d_2+\dots+nd_n}{n}$,but as we know $ d_n=-(d_1+d_2+\dots+d_{n-1})$,so finally $ \boxed{a_1=\frac{d_{n-1}+2d_{n-2}+\dots+(n-1)d_1}{n}}$. Now let's find similar formula for every $ a_k$, where $ 2\leq k\leq n$: $ a_2=a_1-d_1$,$ a_3=a_2-d_2=a_1-d_1-d_2,\dots,\Rightarrow$,$ a_k=a_1-d_1-d_2\dots-d_{k-1}$,combining it with boxed formula,we finally obtain: $ \boxed{a_k=\frac{-d_1-2d_2-\dots-(k-1)d_{k-1}+(n-k)d_k+(n-k-1)d_{k+1}+\dots+d_{n-1}}{n}}$. We need to prove that $ (a_k)^2=\frac{(-d_1-2d_2-\dots-(k-1)d_{k-1}+(n-k)d_k+(n-k-1)d_{k+1}+\dots+d_{n-1})^2}{n^2}\leq\frac{n}{3}(d_1^2+d_2^2+\dots+d_{n-1}^2)$ Using CBS: $ \frac{(-d_1-2d_2-\dots-(k-1)d_{k-1}+(n-k)d_k+(n-k-1)d_{k+1}+\dots+d_{n-1})^2}{n^2}\leq (d_1^2+d_2^2+\dots+d_{n-1}^2)(\frac{1^2+2^2+\dots+(k-1)^2+1^2+2^2+\dots+(n-k)^2}{n^2})$,so it is enough to prove that $ \frac{1^2+2^2+\dots+(k-1)^2+1^2+2^2+\dots+(n-k)^2}{n^2}\leq\frac{n}{3}$,or $ 1^2+2^2+\dots+(k-1)^2+1^2+2^2+\dots+(n-k)^2\leq\frac{n^3}{3}$,but it is easy to understand that: $ 1^2+2^2+\dots+(k-1)^2+1^2+2^2+\dots+(n-k)^2\leq 1^2+2^2+\dots+(n-1)^2=\frac{n(n-1)(2n-1)}{6}\leq\frac{n^3}{3}$. So we are done.

Check my solution, I used convex functions: $ \max_{1\leq i \leq k} a_i^2 \leq \frac{k}{3} \left( (a_1-a_2)^2+(a_2-a_3)^2+\cdots +(a_{k-1}-a_k)^2\right). $ This can be written as a quadratic function depending on $a_i$ for every $1\leq i \leq k$. The first coefficient is positive and as quadratic function s convex, it attains it's minimum at the border meanings of $a_i$. Actually, these borders are $\min_{a_i}$ and $\max_{a_i}$. Denote them as $b$ and $a$ respectively. So, $k-1$ of $a_i$ must equal $a$ or $b$. Let $p$ of the numbers be $a$ and $q$ be $b$. So, the last one equals $-pa-qb$ in order to sum be $0$. And we must prove that $\max_{a^2,c^2} \leq \frac{k}{3}((a-ap+cq)^2+(ap-cq+c)^2)$, where $-b=c$(Note that, $c$ is positive) We can assume that $b^2 \leq a^2$. $c \leq -ap+cq \leq a$ $\frac{q-1}{p} \geq \frac{a}{c} \geq \frac{q}{p+1}$ Now, let's prove that $(a-ap+cq)^2\frac{p+q}{3} \geq a^2$ $(1-p+\frac{c}{a}q)^2 \frac{p+q}{3}\geq (1-p+\frac{pq}{q-1})^2\frac{p+q}{3}=(\frac{p+q-1}{q-1})^2\frac{p+q}{3}\geq 1$ OR: $(p+q-1)^2(p+q)\geq 3(q-1)^2$ Cases when $q=1$ or $0$ are easy. Q.E.D.

IS IT CORRECT OR NOT? CHECK PLEASE =)

Summerburn wrote: How that line comes? I can't see the left side being $\frac{k(k-1)}{3}S$ You're right.That is a mistake in this solution.