 Solution

For an imputation $y$, the excess of a coalition is $e(S,y)=v(S)-{\sum }_{i\in S}{y}_i$. Arrange these excesses in decreasing order. The nucleolus is the imputation that lexicographically minimizes this ordered vector: minimize the largest complaint first, then the next largest among ties, and so on. The empty and grand-coalition excesses are fixed at zero for every efficient allocation, so they can be omitted without changing the lexicographic comparison.

For every efficient $y$ the three pair excesses have sumHence at least one pair has excess at least $-4/3$, and the largest proper-coalition excess is at least $-4/3$. At the proposed allocation all three pair excesses equal $-4/3$, while all three singleton excesses equal $-8/3$. It therefore attains the best possible largest proper-coalition excess.

Moreover, any allocation attaining this bound must have all three pair excesses equal $-4/3$, since they are individually at most that value and sum to $-4$. This forceswhose unique solution is $y=x$. It is already an imputation, so the first nonconstant lexicographic stage has a unique minimizer; there are no later choices to resolve. Therefore