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 all $a_i,b_i,c_i,d_i\leq 1$. Is it always possible to partition $\{1,2,\dots,n\}$ into two subsets $X,Y$ so that at least $3$ of the following $4$ inequalities hold:

$$\left|\sum_Xa_i-\sum_Ya_i\right|\leq 1, \left|\sum_Xb_i-\sum_Yb_i\right|\leq 1$$ $$\left|\sum_Xc_i-\sum_Yc_i\right|\leq 1, \left|\sum_Xd_i-\sum_Yd_i\right|\leq 1$$

An affirmative answer here would imply one for this (unsolved) question.

  • $\begingroup$ As you are asking for "3 out of 4", have you found an example where there is no partition such that all 4 inequalities hold? $\endgroup$ – Wolfgang May 19 '19 at 20:06
  • 1
    $\begingroup$ Yes, for example $(1,1,0,0),(1,0,1,0),(0,1,1,0)$ $\endgroup$ – pi66 May 19 '19 at 21:13

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