Hasin_Ahmad wrote: Let {$a_1, a_2,\cdots,a_n$} be a set of $n$ real numbers whos sym equals S. It is known that each number in the set is less than $\frac{S}{n-1}$. Prove that for any three numbers $a_i$, $a_j$ and $a_k$ in the set, $a_i+a_j>a_k$. I suppose $n\ge 3$. If so : If $a_i+a_j\le a_k$, then $S=\sum a_m=a_i+a_j+a_k+\sum_{m\notin\{i,j,k\}}a_m$ $\le 2a_k+\sum_{m\notin\{i,j,k\}}a_m$ $<2\frac{S}{n-1}+(n-3)\frac S{n-1}=S$, contradiction.

Claim: For any $i,j$, $a_i + a_j >= \frac{S}{n-1}$. Proof: For the sake of contradiction, suppose $a_i + a_j < \frac{S}{n-1}$. Consider a sequence $b_1, b_2, b_3, ..., b_{n-1}$ identical to $a$, but the elements $a_i$ and $a_j$ have been removed but a new element $b_k$ has been inserted with $b_k = a_i + a_j$. This sequence still has the sum $S$, so its average is $A = \frac{S}{n-1}$. So there must exist some $b_i >= A $. But by definition, $b_i < \frac{S}{n-1}$ for all $i$. A contradiction. Then $a_i + a_j >= \frac{S}{n-1} > a_k$.