Order on three real numbers given they are roots of a cubic, their sum and sum of pairwise product

Question

Let $a,b,c$ be three real numbers which are roots of a cubic polynomial and satisfy $a+b+c=6$ and $ab+bc+ca=9.$ Suppose $a<b<c.$ Show that $$0<a<1 <b<3<c<4$$

Source: ISI Bmath Entrance 18-Jul-2021

Let the polynomial be $p(x) = (x-a)(x-b)(x-c)= x^3 - x^2 (6) + 9x - abc$, now clearly $p(a)=p(b)=p(c)$, adding the three:

$$p(a)+p(b)+p(c)=0$$

$$ (a^3 + b^3 + c^3)- 6(a^2 + b^2 +c^2) + 9 (a+b+c) - abc=0$$

Using identites we find the expression above becomes:

$$a^3 + b^3+c^3 - 6(6^2- 2 \cdot 9) + 9 \cdot 6 -abc=0$$

or

$$ a^3 + b^3 + c^3 -54 -abc=0$$

To be frank, I didn't do the manipulatons with any plan in mind, under desperation in the exam hall, I took the above to be a cubic in $a$ and applied Vieta assuming roots to be $\{a_1,a_2,a_3 \}$, where I found:

$$a_1 + a_2 + a_3 = 0$$ $$a_1 a_2 + a_2 a_3 + a_1 a_3 =0$$ $$a_1 a_2 a_3 = b^3 + c^3 - 54 - abc$$

We can write similar relation by considering it as a cubic in $b$ and $c$... but what after this..?

Answer 1

I approached the problem differently. It was a long write-up, here are the key points:

▪︎Define a cubic $f(x)=x^3-6x^2+9x+t$. It is clear that this is the cubic with roots $a,b,c$.

▪︎Try to put a bound on $t$ for which three distinct roots are obtained. The graphical argument can be used. The defined $f(x)$ is merely a shifting of $g(x)=x^3-6x^2+9x$ along the $+y$-axis by $t$ units. If $t\leq-4$ or $t\geq0$, three distinct roots are not obtained. Hence $t\in(-4,0)$.

▪︎We have to find solutions for $x^3-6x^2+9x=k$. Here $k=-t$.We have shown that $k\in(0,4)$.

▪︎To finally solve the problem, we will apply intermediate value theorem upon $g(x)$. Note that $g(0)=0$ and $g(1)=4$. Hence, $g(x)=k$ has a root in $x\in(0,1)$. This is the smallest solution, so $0<a<1$. Also, this interval can have ONLY one solution, since $g$ is monotonic here.

▪︎ Apply the same argument in intervals $(1,3)$ and $(3,4)$. We have finished our proof.

Answer 2

Solution using the derivative:

$P(x)$ is a polynomial, hence its roots are separated by the roots of the first derivative $P'(x).$ $$P'(x)=3x^2-12x+9=3(x-1)(x-3),$$ thus $$a<1<b<3<c.$$

To complete the proof, we have to show that $a>0$ and $c<4.$

1. Let us prove $a>0$ by contradiction.

• Case $a=0:$
  Here, the polynomial is $P(x)=x^3-6x^2+9x=x(x-3)^2,$ which contradicts $b< c.$
• Case $a<0:$
  Now, $abc<0$ because $b,c>1.$ Let us apply Descartes rule of signs. There is only one change of sign in $P(x).$ Therefore, $P(x)$ has exactly one positive root, which contradicts to $\;1<b<3<c\;$ found before.

We conclude that $a>0.$

2. Let us prove that $c<4.$
   From $P'(x)=3(x-1)(x-3)$ we know that $P(x)$ increases on $(-\infty,1)$ and on $(3,\infty)$ and decreases on $(1,3).$ Since $$P(4)=4-abc=P(1)>0,$$ the root $c$ must be less than $4.$