quacksaysduck wrote: Let $n$ be a positive integer and $a_1\le a_2\le\ldots\le a_{n+1}$ and $b_1\le b_2\le\ldots\le b_n$ be real numbers such that for all $k\le n$, \[\binom nk\sum_{\substack{1\le i_1<i_2<\ldots<i_k\le n+1,\\i_1,i_2,\ldots,i_k\in\mathbb N}}a_{i_1}a_{i_2}\ldots a_{i_k} = \binom{n+1}k\sum_{\substack{1\le j_1<j_2<\ldots<j_k\le n,\\j_1,j_2,\ldots,j_k\in\mathbb N}}b_{j_1}b_{j_2}\ldots b_{j_k}.\]Show that \[a_1\le b_1\le a_2\le b_2\le \ldots \le a_n\le b_n\le a_{n+1}.\] (Proposed by Ivan Chan Guan Yu) Let $A=\prod_{i=1}^{n+1}(X+a_i)$, by Rolle theorem, $A'$ hase $n$ real roots $-c_n \leq \cdots \leq -c_1$, such that $a_i \leq c_i \leq a_{i+1}$. So $A'=(n+1)\prod_{i=1}^{n}(X+c_i)$. On the other hand, if we denote $B=\prod_{i=1}^{n}(X+b_i)$, then using properties of symmetric polynomials, the problem statement is equivalent to : $$A'=(n+1)B$$So $\{b_1,...,b_n\}=\{c_1,...,c_n\}$ and hence the result.