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

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

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

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

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