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

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.
