Some $4$-tuples of positive real numbers $(a_1,b_1,c_1,d_1),\dots,(a_n,b_n,c_n,d_n)$ are given, with $$\sum_{i=1}^na_i=\sum_{i=1}^nb_i=\sum_{i=1}^nc_i=\sum_{i=1}^nd_i=3.$$ It is known that there exists a partition of $N=\{1,\dots,n\}$ into three sets $A_1,A_2,A_3$ such that $$\sum_{A_1}a_i=\sum_{A_2}a_i=\sum_{A_3}a_i=1.$$ Analogous statements hold for $b,c,d$. Is it always possible to partition $N$ into two sets $X,Y$ so that $$\sum_X a_i,\sum_X b_i,\sum_Y c_i,\sum_Y d_i\geq 1?$$ When the last line has one $X$ and three $Y$'s, the answer is positive.
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- 5$\begingroup$ You posted this question to m.se, and accepted an answer there, so why are you posting here? math.stackexchange.com/questions/2951883/… $\endgroup$Gerry Myerson– Gerry Myerson2018-10-28 21:16:12 +00:00Commented Oct 28, 2018 at 21:16
- 1$\begingroup$ The question is different. I've clarified. $\endgroup$pi66– pi662018-10-28 21:18:46 +00:00Commented Oct 28, 2018 at 21:18
- $\begingroup$ Seems to be a simpler version of the following (unanswered) question: mathoverflow.net/questions/308839/partition-of-4-tuples $\endgroup$Ilya Bogdanov– Ilya Bogdanov2018-10-30 20:56:09 +00:00Commented Oct 30, 2018 at 20:56
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