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The sum of one hundred given real numbers is zero. Prove that at least 99 of the pairwise sums of these hundred numbers are nonnegative. Is this result the best possible one?

My “solution” is that , we know that not all numbers are negative, otherwise sum would not be zero , thus there is a positive number . Applying the same logic there also exist a negative number.Now there is a sum which equals negative of that negative number . Lets say all numbers are positive except for this negative number , in which case the statement is true . Lets suppose there is at least 1 negative number still. So if we sum those 2 negative numbers we can say that negative of that sum is the remaining numbers . Thus we have 2 sums that are positive .now again either all numbers are positive (which makes the statement true) or there is at least one negative in which case sum of numbers except for the 3negatives that has been established is positive and so on.

m3h3mm3d
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  • There may not be a negative nor a positive number, if all 100 numbers are zero. The statement remains true for this case though. – peterwhy Jun 27 '21 at 16:16
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    I am afraid that you may have misunderstood pairwise sums as subset sums, and there may not be "a sum which equals negative of that negative number". – peterwhy Jun 27 '21 at 16:25
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    It its not true in general that if you have $n$ numbers that sum to zero, then at least $n-1$ pairwise sums are nonnegative. (A counterexample for $n=3$ is ${1, 2, -3}$). So a proof that doesn't appear to have anything in it that makes a difference between $n=100$ and $n=3$ cannot be right. – Troposphere Jun 27 '21 at 16:41
  • It seems i have misunderstood pairwise sum as sum of elements in a subset – m3h3mm3d Jun 27 '21 at 16:51

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It is best possible. Consider the numbers -1, -1, …, -1, 99. These sum to 0, but exactly 99 of the pairwise sums are positive. (If the numbers are required to be distinct, you can easily adjust the solution to satisfy that.)

user432944
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