Prove that following systems of inequalities are equivalent. There are given two systems of inequalities:
$b, d, c, e > 0$ (first system) and $b+d>0, ec>0, (b+d)(e+c+bd)>be+cd, (be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d)^2ec$ (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 $bc+b^2d+de+bd^2>0$ and $bd((e-c)^2+b^2e+bde+bdc+cd^2)>0$ and now it is clear that all inequalities from second system must be true, if $b, c, d, e$ are positive.
The second part is harder and that is all what I found:
we can assume that $b$ is greater or equal than $d$, and from that we know that $b>0$. We know that $e$ and $c$ are both positive or both negative. From the last inequality (from second system) we know that left part is positive, because right is. We know that second factor is positive (it is just transformed third inequality), so first factor is also positive. This factor is right side of third inequality, so we similarly deduce that second factor of third inequality is positive, because right side is positive and first factor is positive. Now we have $e+c+bd>0$ and $be+cd>0$. Assume that $e$ and $c$ are not positive. From $e+c+bd>0$ we know that $d>0$, but this implies that $be+cd<0$. From this contradiction we know that $e, c>0$. Now I need only to show that $d>0$, but I can't find any way to do that. 
Feel free to edit or retag, because I'm not sure where I should post it.
 A: 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, $(be+cd)((b+d)(e+c+bd)-(be+cd))>(b+d)^2ec$, as 
$$\frac{(be+cd)(b+d)}{b^2c^2} bd>\frac{(b+d)^2ec -  (be+cd)((b+d)(e+c)-(be+cd))}{b^2c^2} $$
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 
$e = m c$ and $d = q b$, the LHS has the sign of $q$, and the RHS gives  
$$(1+q)^2m - (m+q)((1+q)(m+1)-(m+q))$$
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 $\leq0$, hence LHS $>0 \geq$  RHS is satisfied. 
Also, the  condition $q > \max (-1, -m)$ is satisfied. 
Further, by the third inequality, also $(1+q)(m+1 +b^2q/c)-(m+q)$ 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 $\geq 0$ so LHS > RHS  can never be attained. Hence the second set of conditions will never produce $q<0$, which is  $d<0$.
$\quad \quad \Box$
