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

Expert

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 $\left(b,d\right)$ and $\left(c,e\right)$, we can assume $b>0$. So it remains to show that $d>0$.
Rewrite the last inequality, $\left(be+cd\right)\left(\left(b+d\right)\left(e+c+bd\right)-\left(be+cd\right)\right)>\left(b+d{\right)}^{2}ec$, 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
$\left(1+q{\right)}^{2}m-\left(m+q\right)\left(\left(1+q\right)\left(m+1\right)-\left(m+q\right)\right)$
Clearly, $m>0$. Further, as is already known, all brackets in here are positive, so $q>max\left(-1,-m\right)$.
The RHS simplifies to
$q\left(-{m}^{2}+2m-1\right)=-q\left(m-1{\right)}^{2}$
Case 1: suppose $q>0$. Then the LHS $>0$ and the RHS $\le 0$, hence LHS $>0\ge$ RHS is satisfied. Also, the condition $q>max\left(-1,-m\right)$ is satisfied. Further, by the third inequality, also $\left(1+q\right)\left(m+1+{b}^{2}q/c\right)-\left(m+q\right)$ must be positive. Since we assume $q>0$, it suffices that $\left(1+q\right)\left(m+1\right)-\left(m+q\right)>0$, which is $mq+1>0$. So from this one, $q>-1/m$, and in total, $q>max\left(-1,-m,-1/m\right)$.
Since either $-m>-1$ or $-1/m>-1$, this reduces to $q>max\left(-m,-1/m\right)$ 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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