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..?
Best Answer
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 $c<4.$
Here, the polynomial is $P(x)=x^3-6x^2+9x=x(x-3)^2,$ which contradicts $b< c.$
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 $a>0.$
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 $4.$