It is obviosly. Sufficiently consider j=1 and $ a_1+a_2+...+a_n=na_1+(a_2-a_1)+..(a_n-a_1)=nb_1+(b_2-b_1)+...+(b_n-b_1)$ give $ a_1=b_1$.

You are too careless, Rust! Can you tell me which is bigger: $ a_{n - 1} - a_{1}$ or $ a_{n} - a_{n - 1}$?

I think I've got a solution, but it's quite messy. Here's a sketch, without all the rigorous details. From the problem statement, it's clear that the sets $ (a_{i} - a_{j}, i > j)$ and $ (b_{i} - b_{j}, i>j)$ coincide. Thus the largest element from each set coincide - i.e. $ a_{n} - a_{1} = b_{n} - b_{1}$, so $ a_{1} - b_{1} = a_{n} - b_{n}$. Since $ a_{1} \leq b_{1}$, we also have $ a_{n} \leq b_{n} ... (1)$. But we also have $ a_{1} + ... + a_{n-1} \leq b_{1} + ... + b_{n-1} ... (2)$. Adding $ (1)$ and $ (2)$, we have $ a_{1} + ... + a_{n} \leq b_{1} + ... + b_{n}$ - however, we have equality here. Thus we have equality both in $ (1)$ and $ (2)$, so $ a_{n} = b_{n}$. Since $ a_{n} - b_{n} = a_{1} - b_{1}$, we also have $ a_{1} = b_{1}$. Lemma: Suppose we have two sequences $ x_{1}, x_{2} ... x_{k}$ and $ y_{1}, y_{2} ... y_{k}$, such that $ x_{i} \geq y_{i}$ for all $ i$, and $ Max [x_{i}] = Max [y_{i}]$. Then for some $ j$, $ x_{j} = y_{j}$, and $ x_{j} \geq x_{i}$, $ y_{j} \geq y_{i}$ for all $ i$. Proof: Suppose $ Max [x_{i}] = x_{u}$, $ Max [y_{i}] = y_{v}$. We have $ x_{u} = y_{v}$. Then $ y_{v} \leq x_{v}$, so $ x_{u} \leq x_{v}$. But $ x_{u} \geq x_{v}$, since $ x_{u}$ is the maximum of set $ x_{i}$. Thus $ x_{u} = x_{v} = y_{v}$. Pick $ j=v$ - now we have $ x_{j} = y_{j}$ and $ x_{j} \geq x_{i}$, $ y_{j} \geq y_{i}$ for all $ i$, as required. Now consider the $ 2$-nd largest element in the set $ (a_{i} - a_{j}, i>j)$ - it is equal to the $ 2$-nd largest element in the set $ (b_{i} - b_{j}, i>j)$. It's not hard to see that this $ 2$-nd largest element is either $ a_{n} - a_{2}$ or $ a_{n-1} - a_{1}$ (since all other elements, except $ a_{n} - a_{1}$, are smaller than one of those two sets). Thus $ Max [a_{n} - a_{2}, a_{n-1} - a_{1}] = Max [b_{n} - b_{2}, b_{n-1} - b_{1}]$. Since $ a_{1} + a_{2} \leq b_{1} + b_{2}$, $ a_{1} = b_{1}$, we have $ a_{2} \leq b_{2}$, so $ a_{n} - a_{2} \geq b_{n} - b_{2}$. Since $ a_{1} + ... + a_{n-1} = b_{1} + ... + b_{n-1}$, and $ a_{1} + ... + a_{n-2} \leq b_{1} + ... + b_{n-2}$, $ a_{n-1} \geq b_{n-1}$, so $ a_{n-1} - a_{1} \geq b_{n-1} - b_{1}$. Now we can apply Lemma, with $ k=2$ to the fact that: $ Max [a_{n} - a_{2}, a_{n-1} - a_{1}] = Max [b_{n} - b_{2}, b_{n-1} - b_{1}]$. From that we have: $ Case 1: a_{n} - a_{2} \geq a_{n-1} - a_{1}$, analogously for $ b_{i}$, and $ a_{2} = b_{2}$; OR $ Case 2: a_{n-1} - a_{1} \geq a_{n} - a_{2}$, analogously for $ b_{i}$, and $ a_{n-1} = b_{n-1}$. Now, using the fact that $ a_{1} = b_{1}$ and $ a_{n} = b_{n}$, we have proved either $ a_{2} = b_{2}$ or $ a_{n-1} = b_{n-1}$. The idea is to split into cases, and keep going. Using the above method, when we have established certain inequalities, and in the case that $ a_{i} = b_{i}$ for $ i = 1,2 ... k$ and $ i = n, n-1 ... n-l$, the above method allows us to deduce that $ a_{k+1} = b_{k+1}$ or $ a_{n-l-1} = b_{n-l-1}$. If we can keep splitting into cases and keep going, by induction, eventually we will have the desired result $ a_{i} = b_{i}$ for all $ i$. So for this strange case-induction to work, we need to prove that the outlined method will always work. I'll post the details of this later.

suppose A={ai - aj | i>j} B={bi - bj | i>j} obviously, A and B coincide so the sum of elements in A(denote as Sa) and in B(denote as Sb) are the same. obviously Sa=(n-1)an+(n-3)an-1+....+(-(n-3)a2)+(-(n-1)a1) Sb=(n-1)bn+(n-3)bn-1+....+(-(n-3)b2)+(-(n-1)b1) since a1+a2+...+an=b1+b2+...+bn so 2n*an+2(n-1)*an-1+...+2k*ak+...+2*a1=Sa+(n+1)(a1+a2+...+an) =Sb+(n+1)(b1+b2+...+bn)=2n*bn+2(n-1)*bn-1+...+2k*bk+...+2*b1 so n*an+...+1*a1=n*bn+...+1*b1 (1) but since a1+..+ai<=b1+..+bi for any 1<=i<=n-1 and a1+...+an=b1+...+bn so ai+...+an>=bi+...+bn for any i (include n) (2) add these n inequality in (2) together we have n*an+...+1*a1>=n*bn+...+1*b1 according to (1), we know that the inequality sign in (2) should all be equal sign which the conclusion follows easily!

The hypothesis tell us that $ \left( a_{1};\;a_{2};\;...;\;a_{n}\right)\prec\left( b_{1};\;b_{2};\;...;\;b_{n}\right)$, so, from Karamata, we know that $ \sum X^{a_i} \geq \sum X^{b_i}$ for every real $ X$. But the hypothesis also tell us that $ (\sum X^{a_i})(\sum X ^{-a_j} ) = (\sum X^{a_i})(\sum X ^{-a_j} )$ so we must have $ \sum X^{a_i} = \sum X^{b_i}$ for every real X, so $ a_i=b_i$ for every i.

I'm sorry for my stupidity But how did you get that The sets of $ (a_i - a_j)$ and $ (b_i - b_j)$ coincde Edited. It's ok now.It seems like i have read only first part of the statement.

mychrom wrote: The hypothesis tell us that $ \left( a_{1};\;a_{2};\;...;\;a_{n}\right)\prec\left( b_{1};\;b_{2};\;...;\;b_{n}\right)$, so, from Karamata, we know that $ \sum X^{a_i} \geq \sum X^{b_i}$ for every real $ X$. But the hypothesis also tell us that $ \sum X^{a_i} \geq \sum X^{b_i}$ for every real $ X$. so we must have $ \sum X^{a_i} = \sum X^{b_i}$ for every real X, so $ a_i=b_i$ for every i. I'm sorry. The function $f(x)=x^{a_i}$ has $f''(x)=a_i(a_i-1)x^{a_i-2} >0$ => f is convex. So I think $ \left( a_{1};\;a_{2};\;...;\;a_{n}\right)\prec\left( b_{1};\;b_{2};\;...;\;b_{n}\right)$ led to $ \sum X^{a_i} \leq \sum X^{b_i}$ for every real $ X$. From $ \sum X^{a_i} \leq \sum X^{b_i}$ (1) for every real $ X$, I don't understand how we can prove that $ \sum X^{a_i} = \sum X^{b_i}$ for every real X. Since $ \left( a_{1};\;a_{2};\;...;\;a_{n}\right)\prec\left( b_{1};\;b_{2};\;...;\;b_{n}\right)$, we conclude that $ \left( -b_{1};\;-b_{2};\;...;\;-b_{n}\right)\prec\left( -a_{1};\;-a_{2};\;...;\;-a_{n}\right)$ => $\sum X^{-a_i} \geq \sum X^{-b_i}$. (2) (1) and (2) doesn't contradict with the condition $ \sum X^{a_i} \sum X^{-a_i} = \sum X^{b_i} \sum X^{-b_i}$ Am I missing something?

I trying find offical solution of USATST 2000-2009 but i can't.Anyone can share it for me?Thanks a lot!

Let $D= \{ d_1, d_2, \dots, d_k \}$ be the set of all real numbers either of the form $a_i - a_j$ or $b_i - b_j$ where $1 \le j < i \le n$. Since the sequences are non-decreasing, it follows that $d_i \ge 0$. Let $k_x$ denote the number of pairs $(i,j)$ where $1 \le j<i \le n$ such that $a_i - a_j=d_x$. As indicated in the problem statement, the number $p_x$ of pairs $(i,j)$ such that $a_i - a_j=d_x$ is equal to the number of pairs $(k,l)$ such that $b_k - b_l = d_x$. However, since $d_x \ge 0$, either $d_x>0$ or $d_x=0$. If $d_x>0$, then each such $(i,j)$ and $(k,l)$ must satisfy that $i>j$. This implies that $p_x=k_x$ and that $k_x$ is also the number of pairs $(k,l)$ where $1 \le l < k \le n$ such that $b_k - b_l = d_x$. If $x$ is such that $d_x=0$, then there must exist exactly $n$ pairs where $i=j$ satisfying $a_i - a_j = 0$ and exactly $n$ pairs where $i=j$ satisfying $b_i -b_j=0$. Furthermore, for each pair $(i,j)$ such that $i>j$ and $a_i - a_j = 0$, the pair $(j,i)$ satisfies $i<j$ and $a_i - a_j = 0$. Hence there is a bijection between the number of pairs $(i,j)$ such that $i>j$ and $a_i - a_j=0$ and the number of pairs $(k,l)$ such that $k<l$ and $a_k - a_l=0$. The same holds for the sequence $\{ b_i \}$. Hence $(p_x - n)/2 = k_x$ and is hence also equal to the number of pairs $(k,l)$ such that $b_k - b_l = 0$. Now consider the sum \[\sum^{k}_{i=1} k_i \cdot d_i = \sum_{1 \le j<i \le n} a_i - a_j = \sum_{1 \le j<i \le n} b_i - b_j\] Regrouping terms in the sums yields that \[\sum^{n}_{i=1} (2i-n-1)a_i = \sum^{n}_{i=1} (2i-n-1)b_i\] which on applying the fact that $a_1 + a_2 + \dots +a_n = b_1 + b_2 + \dots +b_n$ implies that \[\sum^{n}_{i=1} i \cdot a_i = \sum^{n}_{i=1} i \cdot b_i\] However adding the inequalities $a_1 + \dots + a_i \le b_1 + \dots + b_i$ and the equality $a_1 + a_2 + \dots +a_n = b_1 + b_2 + \dots +b_n$ implies that \[\sum^{n}_{i=1} i \cdot a_i \le \sum^{n}_{i=1} i \cdot b_i\] Hence equality holds in every inequality and $S_i = a_1 + \dots +a_i = b_1 + \dots + b_i$ for all $1 \le i \le n$. Hence $S_i - S_{i-1}=a_i=b_i$ for all $2 \le i \le n$ which on implying the equality $a_1 + a_2 + \dots +a_n = b_1 + b_2 + \dots +b_n$ also yields that $a_1 = b_1$.

From the conditions of the problem the following equality holds: \[ (a_1+a_2+\cdots+a_n)^2+\sum_{i<j}(a_i-a_j)^2=(b_1+b_2+\cdots+b_n)^2+\sum_{i<j}(b_i-b_j)^2 \Longrightarrow a_1^2+\cdots+ a_n^2=b_1^2+\cdots+b_n^2 \] But if $a_i \neq b_i$ for some $i$ then $a_1^2+\cdots+ a_n^2>b_1^2+\cdots+b_n^2$ because of majorization reason. Hence $a_i = b_i$ for $i = 1, \cdots, n$.

It is easy to see that $a_i$ majorizes $b_i$ from the givens. Therefore, by Muirhead's inequality we must have \[ \sum_{\text{sym}} x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n} \ge \sum_{\text{sym}} x_1^{b_1}x_2^{b_2} \cdots x_n^{b_n}. \] Furthermore, if $(a_i) \neq (b_i)$ then there are no nontrivial inequality cases (the proof of Muirhead uses weighted AM-GM repeatedly). Let $N = n!+2012$. Taking $(x_1, \cdots, x_n) = (N,\tfrac{1}{N},1,1,1,\cdots,1)$, and view both sides as numbers in base $N$. Each term of the symmetric sum now corresponds to a number of the form $N^a$; there can be no carries since $N \ge n!$ so the condition then implies that each digit is equal. Therefore, the two sides are equal. So this is a nontrivial equality case, which implies the conclusion.

v_Enhance wrote: It is easy to see that $a_i$ majorizes $b_i$ from the givens. Therefore, by Muirhead's inequality we must have \[ \sum_{\text{sym}} x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n} \ge \sum_{\text{sym}} x_1^{b_1}x_2^{b_2} \cdots x_n^{b_n}. \]Furthermore, if $(a_i) \neq (b_i)$ then there are no nontrivial inequality cases (the proof of Muirhead uses weighted AM-GM repeatedly). Let $N = n!+2012$. Taking $(x_1, \cdots, x_n) = (N,\tfrac{1}{N},1,1,1,\cdots,1)$, and view both sides as numbers in base $N$. Each term of the symmetric sum now corresponds to a number of the form $N^a$; there can be no carries since $N \ge n!$ so the condition then implies that each digit is equal. Therefore, the two sides are equal. So this is a nontrivial equality case, which implies the conclusion. The $a_i$'s and $b_i$'s are real, not integers. So how do we write these numbers in base $N$?

@Kiril Bangachev: Yes, you are right, but it's not an essential obstacle. Since multiplying all $a_i,b_i$ by some integer doesn't change the conditions, that idea proves the problem in case $a_i,b_i$ are rationals. After that, arguing by continuity can do the job for the rest of the reals. But that approach would become even more artificial.

@above This doesn't work as you haven't proven $a_n\ge b_n$ v_Enhance wrote: It is easy to see that $a_i$ majorizes $b_i$ from the givens. I think $-a_i$ majorizes $-b_i$

I think this works: We know that $\sum_{i>j} a_i-a_j = \sum_{k>\ell} b_k-b_\ell$ $\sum_{i>j} a_i-a_j = \sum_{i=1}^n (2i-n-1)(a_i) = 2\sum_{i=1}^n ia_i - (n+1)\sum_{i=1}^n a_i$ $\sum_{k>\ell} b_k-b_\ell = \sum_{i=1}^n (2i-n-1)(b_i) = 2\sum_{i=1}^n ib_i - (n+1)\sum_{i=1}^n b_i = 2\sum_{i=1}^n ib_i - (n+1)\sum_{i=1}^n a_i$ $\sum_{i>j} a_i-a_j = \sum_{k>\ell} b_k-b_\ell \rightarrow 2\sum_{i=1}^n ia_i - (n+1)\sum_{i=1}^n a_i = 2\sum_{i=1}^n ib_i -(n+1) \sum_{i=1}^n a_i$ $\rightarrow 2\sum_{i=1}^n ia_i = 2\sum_{i=1}^n ib_i$ So, $\sum_{i=1}^n ia_i = \sum_{i=1}^n ib_i$ By Abel Summation, $\sum_{i=1}^n ia_i= n(a_1+\dots +a_n) + \sum_{i=1}^{n-1} (a_1+ \dots +a_i)(i-(i+1)) = n(b_1+\dots +b_n) - \sum_{i=1}^{n-1} (a_1+ \dots +a_i)$ $\sum_{i=1}^n ib_i= n(b_1+\dots +b_n) + \sum_{i=1}^{n-1} (b_1+ \dots +b_i)(i-(i+1)) = n(b_1+\dots +b_n) - \sum_{i=1}^{n-1} (b_1+ \dots +b_i) $ $\sum_{i=1}^n ia_i = \sum_{i=1}^n ib_i$ $\rightarrow n(b_1+\dots +b_n) - \sum_{i=1}^{n-1} (a_1+ \dots +a_i) = n(b_1+\dots +b_n) - \sum_{i=1}^{n-1}(b_1+ \dots + b_i)$ $\rightarrow - \sum_{i=1}^{n-1} (a_1+ \dots +a_i) = - \sum_{i=1}^{n-1}(b_1+ \dots + b_i)$ $\rightarrow \sum_{i=1}^{n-1} (a_1+ \dots +a_i) = \sum_{i=1}^{n-1}(b_1+ \dots + b_i)$ We know that $a_1+ \dots +a_n = b_1+ \dots +b_n$ So, $ \sum_{i=1}^{n} (a_1+ \dots +a_i) = \sum_{i=1}^{n}(b_1+ \dots + b_i)$ Thus, $a_i = b_i$ for $i=1, \dots, n$. $\blacksquare$

By the given condition, $$\sum_{1 \leq i < j \leq n}(a_i - a_j) = \sum_{1 \leq i < j \leq n}(b_i - b_j),$$as both sides only sum over all of the nonnegative differences of the form $a_i - a_j$ and $b_k - b_l$. Combining like terms gives us $$\sum_{i=1}^n(n-2i+1)(a_i) = \sum_{i=1}^n(n-2i+1)(b_i),$$which can be rewritten as $$(1-n)\sum_{i=1}^n a_i + 2\sum_{i=1}^{n-1}(a_1+a_2+\dots+a_i) = (1-n)\sum_{i=1}^n b_i + 2\sum_{i=1}^{n-1}(b_1+b_2+\dots+b_i).$$It is given that $a_1 + \dots + a_n = b_1 + \dots + b_n$, so we can cancel the first term of each side to get $$\sum_{i=1}^{n-1}(a_1+a_2+\dots+a_i) = \sum_{i=1}^{n-1}(b_1+b_2+\dots+b_i).$$Note that it is given that $a_1 + \dots + a_i \le b_1 + \dots + b_i$ for every integer $i$ in $[1, n-1]$. As summing together all of these conditions gives our derived equality (above), all of these inequalities must be equalities, and the desired result follows. $\square$

Abel Summation is enough