Prove that following systems of inequalities are equivalent.There are given two systems of inequalities: b...

antennense

antennense

Answered

2022-06-29

Prove that following systems of inequalities are equivalent.
There are given two systems of inequalities: b , d , c , e > 0 (first system) and b + d > 0 , e c > 0 , ( b + d ) ( e + c + b d ) > b e + c d , ( b e + c d ) ( ( b + d ) ( e + c + b d ) ( b e + c d ) ) > ( b + d ) 2 e c (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 b c + b 2 d + d e + b d 2 > 0 and b d ( ( e c ) 2 + b 2 e + b d e + b d c + c d 2 ) > 0 and now it is clear that all inequalities from second system must be true, if b , c , d , e are positive.

Answer & Explanation

Tamia Padilla

Tamia Padilla

Expert

2022-06-30Added 16 answers

OP has already shown that c , e > 0 and that one of b,d must be > 0. Since the inequalities are all unchanged under simultaneous exchange of ( b , d ) and ( c , e ), we can assume b > 0. So it remains to show that d > 0.
Rewrite the last inequality, ( b e + c d ) ( ( b + d ) ( e + c + b d ) ( b e + c d ) ) > ( b + d ) 2 e c, as
On the LHS all factors "in front" of b d are > 0, hence the sign of the LHS will be determined by the sign of b d.
Writing e = m c and d = q b, the LHS has the sign of q, and the RHS gives
( 1 + q ) 2 m ( m + q ) ( ( 1 + q ) ( m + 1 ) ( m + q ) )
Clearly, m > 0. Further, as is already known, all brackets in here are positive, so q > max ( 1 , m ).
The RHS simplifies to
q ( m 2 + 2 m 1 ) = q ( m 1 ) 2
Case 1: suppose q > 0. Then the LHS > 0 and the RHS 0, hence LHS > 0 RHS is satisfied. Also, the condition q > max ( 1 , m ) is satisfied. Further, by the third inequality, also ( 1 + q ) ( m + 1 + b 2 q / c ) ( m + q ) must be positive. Since we assume q > 0, it suffices that ( 1 + q ) ( m + 1 ) ( m + q ) > 0, which is m q + 1 > 0. So from this one, q > 1 / m, and in total, q > max ( 1 , m , 1 / m ).
Since either m > 1 or 1 / m > 1, this reduces to q > max ( m , 1 / m ) which is satisfied. Hence solutions of the second set of conditions with q > 0 exist, which implies d > 0.
Case 2: suppose q < 0. Then LHS < 0 and the RHS 0 so LHS > RHS can never be attained. Hence the second set of conditions will never produce q < 0, which is d < 0.

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