Prove that following systems of inequalities are equivalent.
There are given two systems of inequal
Prove that following systems of inequalities are equivalent. There are given two systems of inequalities: (first system) and (second system). I need to show that they are equivalent. It is easy to see that first system implies second, because last two inequalities (from the second one) can be transformed to form and and now it is clear that all inequalities from second system must be true, if are positive.
Answer & Explanation
Beginner2022-06-30Added 16 answers
OP has already shown that and that one of b,d must be >. Since the inequalities are all unchanged under simultaneous exchange of and , we can assume . So it remains to show that . Rewrite the last inequality, , as On the LHS all factors "in front" of are >, hence the sign of the LHS will be determined by the sign of . Writing and , the LHS has the sign of q, and the RHS gives
Clearly, . Further, as is already known, all brackets in here are positive, so . The RHS simplifies to
Case 1: suppose . Then the LHS and the RHS , hence LHS RHS is satisfied. Also, the condition is satisfied. Further, by the third inequality, also must be positive. Since we assume , it suffices that , which is . So from this one, , and in total, . Since either or , this reduces to which is satisfied. Hence solutions of the second set of conditions with exist, which implies . Case 2: suppose . Then LHS and the RHS so LHS > RHS can never be attained. Hence the second set of conditions will never produce , which is .
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