# 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.

• It is not clear what is the first system and what is the second Commented Apr 18, 2017 at 11:59
• By the first system I mean that all numbers (b, c, d, e) are positive. The second system is the rest of inequalities. I distinguished between them by putting "and" between. Anyway, thanks, I will edit to make it more clear. Commented Apr 18, 2017 at 12:13
• I agree that, without loss of generality, $b > 0$ and it follows that $c,e>0$. But then I think the second system is valid with $d=0$, provided $c < e$. For example, $b = c = 1$, $d = 0$ and $e = 2$. Perhaps also for some value of $d < 0$, but in this case, it should be $d > -bc/e$. Commented Apr 18, 2017 at 14:01
• I think fourth equation of this system is not valid for this values. I think I have mistaken fourth inequality in my post (extremely sorry for that, I hope that now everything is right), but all computations are right (I checked my computations second time). This problem is just reformulation of Routh-Hurtwitz criterion for stability of a polynomial, which I wanted to "check-by-hand" for polynomials of degree 4, so I think it should be correct. Commented Apr 18, 2017 at 14:19

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 $d<0$.

$\quad \quad \Box$

• "q>max(−1,−m)" This is assuming (for contradiction) that q<0? I will be very grateful if you explain how can I get this "(1+q)(m+1)−(m+q)>0" from the third inequality (probably I am making the same computational mistake time by time). Commented Apr 21, 2017 at 22:10
• "q>max(−1,−m)" assumes nothing about q (the assumptions about the sign of q come at the bottom). It is just the formulation (in q and m) of the statements that you already observed, $b+d>0$ and $be + cd > 0$. As to your other question: the problem's third inequality is not correctly rephrased, I will change that in the text. Commented Apr 21, 2017 at 22:18
• The third inequality is now correctly discussed in the text. Commented Apr 21, 2017 at 22:26
• Now everything is clear. Thanks for your elaboration, great solution! (+1) Commented Apr 23, 2017 at 15:07