Necessary and sufficient conditions that a cubic equation has three positive real roots

Question

Given the cubic equation $$x^3+px^2+qx+r=0$$

What are the necessary and sufficient conditions that this equation has three positive real roots?

My attempt:

From this answer, the necessary and sufficient conditions that a cubic equation has three real roots is $$-27r^2 + 18 pqr - 4 q^3 - 4 p^3 r + p^2 q^2 \ge 0 \tag{1}$$

In order to make the roots positive, the necessary conditions are $$p <0\tag{2}$$ $$q >0\tag{3}$$ $$r <0\tag{4}$$ but are these conditions sufficient?

PS: Finally, I found the answer.

Answer 1

if a cubic should have 3 positive real root, lets say the roots are $a$, $b$ and $c$ then the cubic equation can be written as $(x-a)\cdot(x-b)\cdot(x-c)$, now if you expand it $$x^3-(a+b+c)\cdot x^2+(a.b+a.c+b.c)\cdot x-a.b.c$$ if truly $a ,b, c$ are positive real, then coeffient of $x^2$ is $ < 0$, coefficient of $x $ is $> 0 $ and coefficient of $x^0 $ is $< 0 $ as you have clearly stated, yes it's the sufficient conidition

Answer 2

Let $a,b,c$ be three roots of the cubic equations. We will prove that if $(2),(3)$ and $(4)$ hold, $a,b,c$ must be all positive.

Suppose the contradiction, from $(4)$ we can suppose that $a<0,b<0$ and $c>0$.

From $(2)$, we have $c> -a-b$.

From $(3)$, we have $$ab + c(a+b)>0 \implies ab >c(-a-b)>(-a-b)^2 = a^2 +2ab+b^2$$ or $a^2+ab+b^2 <0$ (contradiction).

So, $a,b,c$ must be all positive.

Then, $(1),(2),(3)$ and $(4)$ are necessary and sufficient conditions for having all 3 positive roots.