3

For $a,b,c\in\mathbb{R}_+$, let $\sum a^3=x,\sum a^2(b+c)=y,abc=z$

By AM-GM and Schur inequality, we get

$6z\le y\le x+3z$.

But I wonder whether it is the sufficient condition. That is, given $x,y,z\in \mathbb{R}_+,6z\le y\le x+3z$, if there are $a,b,c\in \mathbb{R}_+,\sum a^3=x,\sum a^2(b+c)=y,abc=z$?

My try:

Let $$(a+b+c)^3=x+3y+6z=-B^3$$ $$(a+b+c)(ab+ac+bc)=y+3z=-B\cdot C$$ $$abc=z=-D$$

$a,b,c$ are three real roots of equation $X^3+BX^2+CX+D=0$

We can use Cardano discriminant to find if the equation always has three real roots. But it seems too complicated. I want a conciser one.

Appreciate your help.

Kurippu
  • 93

1 Answers1

2

Hint:

The cubic equation $u^3 + Bu^2 + Cu + D = 0$ has three non-negative real roots if any only if $B \le 0, C \ge 0, D \le 0$ and its discriminant is non-negative, i.e. $$\Delta := -4B^3D + B^2C^2 + 18BCD - 4C^3 - 27D^2 \ge 0. \tag{1}$$ (The proof is similar to Necessary and sufficient conditions that a cubic equation has three positive real roots)

The condition is $x, y, z \in \mathbb{R}_{\ge 0}$ and $6z\le y\le x+3z$ and (1).

Without (1), it is not true. For example, $x = 9, y = 12, z = 1$.

River Li
  • 37,323
  • Thanks for your help. – Kurippu Mar 21 '22 at 02:10
  • @Kurippu You are welcome. – River Li Mar 21 '22 at 02:24
  • @RiverLi How do you show this fact, that the cubic has three non-negative real roots iff all those inequalities hold ? The "only if" is clear to me. In the other direction, from $\Delta \geq 0$ we deduce that the cubic has three real roots, but I don't see yet how to show that all those roots are non-negative. – Ewan Delanoy Mar 21 '22 at 11:45
  • 1
    @EwanDelanoy It is similar to https://math.stackexchange.com/questions/4058814/necessary-and-sufficient-conditions-that-a-cubic-equation-has-three-positive-rea. If you understood that, you consider the remaining cases $B = 0$ or $C = 0$ or $D = 0$. – River Li Mar 21 '22 at 12:36
  • @RiverLi Got it, thanks. – Ewan Delanoy Mar 21 '22 at 13:04