Suppose the contrary. Replace all numbers $a_{i+1}, ..., a_n$ with $k\frac{a_{i+1}+...+a_n}{n-i}$, where $k>1$ such number that $a_{i+1}^2+...+a_n^2=(n-i)k^2\left(\frac{a_{i+1}+...+a_n}{n-i}\right)^2$. We see that $m$ is increases for the new set, so $(n-i)\geq n(m-a_i)^2$ is still wrong. Let's increse $a_{i-1}$ and decrease $a_{i+1}$ preserving $a_{i-1}^2+a_{i+1}^2$. Then $a_{i-1}+a_{i+1}$ increases (since $a_{i-1}\leq a_i\leq a_{i+1}$. Finally we will get $a_{i-1}=a_{i}$ or $a_i=a_{i+1}$. Average again $a_{i+1},...,a_n$. Perform this operation many-many times (maybe infinitely many - doesn't matter, since $(m-a_i)$ non-decrease at each step). Finally we will necessary obtain the set of $i$ numbers, which equals to $a_i=a$ and $n-i$ numbers which equals to $a_{i+1}=b$, $a\leq b$. Thus we have $m=\frac{ia+(n-i)b}{n}$, $ia^2+(n-i)b^2=n$, moreover we know that $n-i<n(m-a)^2$. It means that $n<(n-i)(b-a)^2$, so \[ia^2+(n-i)b^2<(n-i)(b-a)^2.\] Therefore $b<\frac{n-2i}{2(n-i)}a<a$ -- contradiction. I hope it is correct.

Well, nice solution Myth. But the problem is that I understood nothing from what you wrote . Don't you have something in High school level ? Or maybe Omid can post the official solution

I don't know the official solution but this is my solution at the exam. Consider $b_k$ like this $b_k=m-a_k$. Therefore we have $b_1 \geq b_2 \geq \dots \geq b_n$. Because $a_k$ are positive then $b_k \leq m$. Now : \[\sum_{i=1}^{n} b_i=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{n} b_i^2 =n(1-m^2)\] Now we have $b_i \geq 0$ and we must prove that: $b_i^2 \leq {n-i \over n}$. First you see: \[b_1+\dots+b_n \geq ib_i\] So \[b_{i+1}+\dots+b_n \leq -ib_i\] From Cauchy-Shwartz inequality we have: \[ b_1^2+\dots+b_n^2 \geq {(b_1+\dots+b_i^2) \over i} \ \ \ \ \ \ \ \ \ \ \ b_{i+1}^2+\dots+b_n^2 \geq {(b_{i+1}+\dots+b_n^2) \over {n-i}}\] And adding the two inequalities: \[n(1-m^2)= \displaystyle \sum_{i=1}^{n} b_i^2 \geq {{i.n.b_i^2}\over{n-i}}\] So \[ b_i^2 \leq {{(n-i)(1-m^2)}\over{i}}\] Now we know $b_i^2 \leq m^2$ because of the problem condition. So \[b_i^2 \leq min \ \Big \{m^2,{{(n-i)(1-m^2)}\over{i}} \Big \} \leq {n-i \over n}\] The last inequality is obvious and we're done.

official solution is really easy!!! $ {(\sum_{j=1}^{i} a_j)\over n}+{(\sum_{j=i+1}^{n} a_j)\over n}=m\leq a_i+{(\sqrt{(\sum_{j=i+1}^{n}a_j^2)(n-i)})\over n}\leq a_i +{(\sqrt{n-i})\over (\sqrt{n})} $ ,it is finished

Wow! I wonder who managed to find this solution.

By replacing $a_1,a_2...a_i,a_n$ with $0,0,0...0, \sqrt{a_1^2+a_2^2+....+a_i^2+a_n^2}$ we get $m'-a_i'=m'>m-a_i$, where $m',a_i'=0$ are the new values, so now we only have to solve it in the case when $a_1=0,a_2=0...a_i=0$ which is easy.

Call $b_i=m-a_i\implies \sum b_i=0,\sum b^2_i=n(1-m^2)$ now suppose $b_1\geq b_2\geq b_3......b_i>0>b_{i+1}\geq ....\geq b_n$. Now just by CS we've $n\geq\sum b^2_i\geq \frac {n}{i(n-i)}\times \sum_{k=1}^{i} b^2_k$ ,and so we get $(n-i)i\geq n^2b^2_i$ ,therefore we're done.