antennense

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 $b+d>0,ec>0,(b+d)(e+c+bd)>be+cd,(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d{)}^{2}ec$ (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 $bc+{b}^{2}d+de+b{d}^{2}>0$ and $bd((e-c{)}^{2}+{b}^{2}e+bde+bdc+c{d}^{2})>0$ and now it is clear that all inequalities from second system must be true, if $b,c,d,e$ are positive.

There are given two systems of inequalities: $b,d,c,e>0$ (first system) and $b+d>0,ec>0,(b+d)(e+c+bd)>be+cd,(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d{)}^{2}ec$ (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 $bc+{b}^{2}d+de+b{d}^{2}>0$ and $bd((e-c{)}^{2}+{b}^{2}e+bde+bdc+c{d}^{2})>0$ and now it is clear that all inequalities from second system must be true, if $b,c,d,e$ are positive.

Tamia Padilla

Beginner2022-06-30Added 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, $(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d{)}^{2}ec$, as

On the LHS all factors "in front" of $bd$ are >$0$, hence the sign of the LHS will be determined by the sign of $bd$.

Writing $e=mc$ and $d=qb$, 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

$q(-{m}^{2}+2m-1)=-q(m-1{)}^{2}$

Case 1: suppose $q>0$. Then the LHS $>0$ and the RHS $\le 0$, hence LHS $>0\ge $ 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 $mq+1>0$. 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 $\ge 0$ so LHS > RHS can never be attained. Hence the second set of conditions will never produce $q<0$, which is $d<0$.

Rewrite the last inequality, $(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d{)}^{2}ec$, as

On the LHS all factors "in front" of $bd$ are >$0$, hence the sign of the LHS will be determined by the sign of $bd$.

Writing $e=mc$ and $d=qb$, 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

$q(-{m}^{2}+2m-1)=-q(m-1{)}^{2}$

Case 1: suppose $q>0$. Then the LHS $>0$ and the RHS $\le 0$, hence LHS $>0\ge $ 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 $mq+1>0$. 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 $\ge 0$ 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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