Without relating the problem to polynomials, how can we solve this inequality? $a,b,c$ are real;
$a<b<c$ ;
$a+b+c=6$ ;
$ab+bc+ca = 9$.
Prove that $0<a<1<b<3<c<4$.
I solved this by treating $a,b,c$ as roots of $x^3 - 6x^2 + 9x + d$ for some $d$, but I have not been able to solve it in a more direct way.

Partial solution:
Let $\{x,y,z\} =\{a,b,c\}$. Since $z=6-x-y$ we get a quadratic equation:
$$ x^2+x(y-6)+(y-3)^2 =0$$ 
and since this one has to have a real solution we have a discriminant nonnegative, so
$$ -3y(y-4)\geq 0 \implies y\in [0,4]$$ 
and the same we can say for $x$ and $z$. 

Now can any of $x,y,z\in\{0,4\}$? 
Say $z=0$ then we get $$6=x+y >2\sqrt{xy} = 2\cdot 3 =6$$ 
a contradiction. 
Say $z=4$ then we get $x+y=2$ and so $xy=1$ so $$2=x+y >2\sqrt{xy} = 2\cdot 1 =2$$ 
a contradiction again. 

So if we return to $a,b,c$ we get: $0<a<b<c<4$.
Now if $b\geq 3$ then $c>3$ and thus $a+b+c>6$ a contradiction.
If $b\leq 1$ then we get $ a<1$ and so $a+b+c<6$ a contradiction again. So $$1<b<3$$ 
We are left to prove $a<1$ and $c>3$....
