Where are the other problems?

I added the problems to the contests section, you can see them here. Thanks zhaobin for posting the problems

Could you post the official solution or other ?

Restore...

Let $P(n)$ be "$\forall a_1,a_2,\ldots,a_n \in R , \sum_{i=1}^na_i^2-\sum_{i=1}^n a_i a_{i+1} \le \lfloor \frac{n}{2}\rfloor(M-m)^2.$ where $ a_{n+1}=a_1,M=\max_{1\le i\le n} a_i,m=\min_{1\le i\le n} a_i.$" obviously, $P(1)$ and $P(2)$ are true. if $P(2m-1)$ is true. i.e.$\forall a_1,a_2,\ldots,a_{2m-1}\in R$ , $(a_1^2+a_2^2+...+a_{2m-1}^2)-(a_1a_2+a_2a_3+...+a_{2m-1}a_1) \le (m-1)(\max_{1\le i\le 2m-1} a_i-\min_{1\le i\le 2m-1} a_i)^2.$ for $n=2m$ , $(a_1^2+a_2^2+...+a_{n+1}^2)-(a_1a_2+a_2a_3+...+a_na_{m+1}+a_{n+1}a_1) \\\ =(a_1^2+a_2^2+...+a_{n}^2)-(a_1a_2+...+a_{n-1}a_{n}+a_na_1)+(a_{n+1}-a_n)(a_{n+1}-a_1).....(*)$ $\forall a_1,a_2,\ldots,a_{2m}\in R$, WLOG$ \max_{1\le i\le 2m} a_i,\min_{1\le i\le 2m} a_i \in \{a_1,a_2,...,a_{2m-1}\}$, take $n=2m-1 $in $(*)$, $(a_1^2+a_2^2+...+a_{2m}^2)-(a_1a_2+a_2a_3+...+a_{2m-1}a_{2m}+a_{2m}a_1) \\\ =(a_1^2+a_2^2+...+a_{2m-1}^2)-(a_1a_2+...+a_{2m-2}a_{2m-1}+a_{2m-1}a_1)+(a_{2m}-a_{2m-1})(a_{2m}-a_1) \\\ \le (m-1)(\max_{1\le i\le 2m-1} a_i-\min_{1\le i\le 2m-1} a_i)^2+(\max_{1\le i\le 2m} a_i-\min_{1\le i\le 2m} a_i)^2 \\\ \le m(\max_{1\le i\le 2m} a_i-\min_{1\le i\le 2m} a_i)^2$ therefore $P(2m)$ is true. If$ P(2m)$ is true. $\forall a_1,a_2,\ldots,a_{2m+1} \in R$, (let $a_{2m+1+t}=a_t$) if there is a $a_i$ s.t. $(a_i-a_{i+1})(a_i-a_{i-1} )\le 0$ and $ \max_{1\le i\le 2m+1} a_i,\min_{1\le i\le 2m+1} a_i \in \{a_1,a_2,...,a_{i-1},a_{i+1},...,a_{2m}\}$, then we are done since we can use$P(2m)$ and (*) on $a_{i+1},a_{i+2},...,a_{2m+1},a_{1},...,a_{i-1} $ if there is a $a_i$ s.t. $a_i=a_{i+1}$ , we can use (*) and P(2m) ,then we are done. suppose there is no $a_i$ s.t. '''$(a_i-a_{i+1})(a_i-a_{i-1}) \le 0$ and $ \max_{1\le i\le 2m+1} a_i,\min_{1\le i\le 2m+1} a_i \in \{a_1,a_2,...,a_{i-1},a_{i+1},...,a_{2m}\}$''' and no $i$ s.t. '$a_i=a_{i+1}$': WLOG suppose $a_1<a_2$, then $a_2>a_3,a_3<a_4,...,a_{2m-1} <a_{2m},a_{2m}>a_{2m+1}.$ since (a_i-a_{i+1})(a_i-a_{i-1})>0 if $a_i=\max_{1\le i\le 2m+1} a_i$ or $\min_{1\le i\le 2m+1} a_i$, we find that a_{2m+1}<a_1 or a_{2m+1}>a_1 are impossible. $\therefore P(2m+1)$ is true. By MI, $P(n)$ is true

Could anyone post a soln i can read? i need to teach my stus.

thesamyaandoctor wrote: Could anyone post a soln i can read? i need to teach my stus. which part you can understand?? I think I can clarify it

Inequality is equivalent to $\sum_{i=1}^{n}(a_{i} - a_{i+1})^2\le\left\lfloor 2\frac{n}{2}\right\rfloor(M-m)^{2}$. For $n$ even, this is obvious: each term is less than $(M-m)^2$. For $n$ odd, fix $m,M$ and use the fact that the left-hand side is a convex function of $a_i$ for all $i$ and so its maximal value is attained at $a_i = m$ or $M$ for all $i$, observe this and the inequality is trivial.

∫FaiLurE∮ wrote:

How can you assume that $ \max_{1\le i\le 2m} a_i,\min_{1\le i\le 2m} a_i \in \{a_1,a_2,...,a_{2m-1}\}$

From $ (a_i - m)(a_i - M)\le 0$ we get ${a_i}^2 + Mm \le a_i(M+m)$ and then $\sum_{i=n}^n {a_i}^2 + nMm \le (M+m)\sum_{i=n}^n a_i$ (1) Again, from $ 0 \le (a_i - M)(a_{i+1} - M) $ and $0 \le (a_i - m)(a_{i+1} - m) $ we get $ -n(m^2 + M^ 2) + 2(M+m)\sum_{i+1}^n a_i \le 2\sum_{i=1}^n a_i a_{i+1} $ (2) Summing 2(1) + (2) we are done.

Gabriel2T wrote: From $ (a_i - m)(a_i - M)\le 0$ we get ${a_i}^2 + Mm \le a_i(M+m)$ and then $\sum_{i=n}^n {a_i}^2 + nMm \le (M+m)\sum_{i=n}^n a_i$ (1) Again, from $ 0 \le (a_i - M)(a_{i+1} - M) $ and $0 \le (a_i - m)(a_{i+1} - m) $ we get $ -n(m^2 + M^ 2) + 2(M+m)\sum_{i+1}^n a_i \le 2\sum_{i=1}^n a_i a_{i+1} $ (2) Summing 2(1) + (2) we are done. It just get a proof for $\frac{n}{2}$ in the right, not $\lfloor{\frac{n}{2}}\rfloor$. It's simple to show that if n is odd, there exists $a_{i-1}, a_i, a_{i+1}$ satisfying $(a_{i-1} - a_i)^2 + (a_i - a_{i-1})^2 \leq (M-m)^2$ and complete the proof.

Part $I$: For even $n$, this holds. Proof of Part $I$: Let $n=2k$. \[(a_1^2+\dots+a_{2k}^2)-(a_1a_2+\dots+a_{2k}a_1)\overset{?}{\leq} k(M-m)^2\iff \sum{(a_i-a_{i-1})^2}\overset{?}{\leq} 2k(M-m)^2\]Which is true since $(M-m)^2\geq (a_i-a_{i-1})^2$. Part $II$: For odd $n$, it holds. Proof of Part $II$: Induct on $n$. It's true for evens. Let $n$ be odd. Denote by $f(n)=RHS-LHS$. Note that for odd $n$, there exists $(a_i-a_{i-1})(a_i-a_{i+1})\leq 0$. Choose $a_n$ as $(a_n-a_{n-1})(a_n-a_1)\leq 0$. \[f(n)-f(n-1)=\sum_{i=1}^{n-1}{a_i^2}-\sum_{i=1}^{n}{a_i^2}+a_{n-1}a_n+a_na_1-a_{n-1}a_1=(a_n-a_{n-1})(a_1-a_n)\geq 0\]As desired.$\blacksquare$