Suppose, one hundred real numbers are given and their sum is 0.Then how can I prove that at least 99 of the pairwise sums of these hundred numbers are non-negative?
I tried this: Let the real numbers be $a_i$,$1\leq i\leq100$. By the given condition, $a_1+a_2+\dots +a_{100}=0$.Suppose for the sake of contradiction, that at most 98 of the pairwise sums of these given reals are non-negative.I just tried to examine the case when exactly 98 of them are non-negative. I labelled the sums $S_1$,$S_2,\dots S_{4950}$,WLOG assume that sums $S_1,S_2,\dots S_{4852}$ are negative, the rest are non-negative. $S_1+S_2+\dots +S_{4950}=99(\sum_{i=1}^na_i)=0$ Case I: $S_{4853}+a_{4854}+\dots +a_{4950}=0$.That simply means that $S_1+S_2+\dots +S_{4852}=0$.But, by the given condition, $S_1+S_2+\dots +S_{4852}<0$,a contradiction.After that, I am not able to make any progress.
Can anyone please suggest a more efficient method to solve this problem.I feel this approach will involve enormous case work.