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If there exists a solution for following system of inequalities, find at least one, else show that there is no solution exists.

$$x_4(x_1+x_7)>x_3(x_2+x_8)$$ $$x_3(x_5+x_8)>x_7(x_4+x_1)$$ $$x_7(x_6+x_4)>x_8(x_5+x_3)$$ $$x_8(x_3+x_2)>x_4(x_6+x_7)$$

for all $i$, $0 < x_i <1$ and $\sum_i x_i=1$

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1 Answer 1

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A quick computer simulation shows such solutions exist. Example:

A = (0.020302010304655, 0.052938093317538, 0.040635876493691, 0.016599197409892, 0.247812843241618, 0.120692838614077, 0.388175655457951, 0.112843485160578)

sum(A)

ans = 1

$A(4)*(A(1)+A(7))-A(3)*(A(2)+A(8))$

ans = 4.372166355951553e-05

$A(3)*(A(5)+A(8))-A(7)*(A(4)+A(1))$

ans = 3.314355258354201e-04

$A(7)*(A(6)+A(4))-A(8)*(A(5)+A(3))$

ans = 0.020743867247722

$A(8)*(A(3)+A(2))-A(4)*(A(6)+A(7))$

ans = 0.002112404285034

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  • $\begingroup$ @ pharmine: Converting the elements of $A$ to rationals and adding those, one obtains $${\frac {112204726358063191347750231121831481}{ 112204726389776842878359836410752160}}, $$ not $1$. $\endgroup$
    – user64494
    Commented Feb 4, 2015 at 12:53
  • $\begingroup$ @user64494 I'll check, but I think that can be dealt with without affecting inequality signs. $\endgroup$
    – pharmine
    Commented Feb 4, 2015 at 14:36
  • $\begingroup$ Thank you, I changed $A(8)$ to satisfy $A(1)+\cdots+A(8)=1$ and updated the results accordingly. $\endgroup$
    – pharmine
    Commented Feb 4, 2015 at 14:41
  • $\begingroup$ @ pharmine: For "updated values" one obtains the same $$ {\frac {112204726358063191347750231121831481}{ 112204726389776842878359836410752160}} .$$ $\endgroup$
    – user64494
    Commented Feb 4, 2015 at 15:04
  • $\begingroup$ Really? Even though $0.112843485160579$ was changed to $0.112843485160578$? You might need to refresh the page. $\endgroup$
    – pharmine
    Commented Feb 4, 2015 at 15:06

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