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.
 A: I think you are going in the right direction. You need some estimates of the coefficient $d$. Note that the polynomial 
$$P(x)=x^3 - 6x^2 + 9x + d=x(x-3)^2+d$$ 
has three real distinct roots $a<b<c$ if and only if $P(1)=4+d>0$ and $P(3)=d<0$ where $1$ is the local maximum of $P$ and and $3$ is local minimum of $P$ (check the derivative $P'$). Therefore $-4<d<0$. 
Now consider the signs of the values $P(0)$, $P(1)$, $P(3)$, and $P(4)$.
What may we conclude?
Batominovski's Comment:  Your answer, albeit not what the OP wants, is useful.  Please consider putting it back, so that someone can benefit from it.
A: I want to note that solving this problem using polynomials may be the most natural and, in fact, direct way to do.  You can for example prove each inequality individually.  To show $a>0$, you have $$0 < (b-c)^2=(b+c)^2-4bc=(6-a)^2-4\big(9-a(b+c)\big)=(6-a)^2-4\big(9-a(6-a)\big)\,,$$
which gives $3a(4-a)>0$, whence $a>0$ (and similarly, $c<4$).
Now, since $0<a<b<c$, we have
$$
\begin{align}0<(b-a)(c-a)&=bc-(b+c)a+a^2=\big(9-(b+c)a\big)-(b+c)a+a^2
\\&=9-2(b+c)a+a^2=9-2(6-a)a+a^2=3(a-1)(a-3)\,.
\end{align}$$
That is, either $a<1$ or $a>3$.  Clearly, $a<\frac{a+b+c}{3}=2$, so $a<1$ must hold.  In the same way, a similar proof shows that $c>3$.
Finally, note that
$$\begin{align}
0<(b-a)(c-b)=-ac+(a+c)b-b^2=-9+2(a+c)b-b^2=-9+2(6-b)b-b^2\,.
\end{align}$$
This shows that $0<3(b-1)(3-b)$, or $1<b<3$.  To be honest, I find this proof rather cumbersome and inelegant.

In general, suppose that $p(x)=(x-a)(x-b)(x-c)=x^3-ux^2+vx-w$, where $a,b,c,u,v,w\in\mathbb{R}$ with $a\leq b\leq c$.  By observing that $$p'(x)=3\left(x^2-\frac{2}3ux+\frac13v\right)=3\left(x-\frac{1}{3}u+\frac{1}{3}\Delta\right)\left(x-\frac{1}{3}u-\frac{1}{3}\Delta\right)\,,$$
where $\Delta:=\sqrt{u^2-3v}$ (i.e., $u^2\geq 3v$ must hold), we conclude that $$a\leq \frac{u-\Delta}{3}\leq b\leq \frac{u+\Delta}{3} \leq c\,.$$
The minimum possible value of $a$ occurs when $w$ is minimized, which happens when $b=c=\frac{u+\Delta}{3}$, and so $a=u-b-c=\frac{u-2\Delta}{3}$.  Consequently,
$$a\geq \frac{u-2\Delta}{3}$$
must hold in general.  Similarly, the maximum possible value of $c$ occurs when $w$ is maximized, so that $a=b=\frac{u-2\Delta}{3}$ and $c=u-b-c=\frac{u+2\Delta}{3}$.  This means that
$$c\leq \frac{u+2\Delta}{3}$$ in general.  In other words, if real numbers $a,b,c$ satisfy $a\leq b\leq c$, $a+b+c=u$, and $bc+ca+ab=v$, then
$$\frac{u-2\sqrt{u^2-3v}}{3}\leq a \leq \frac{u-\sqrt{u^2-3v}}{3}\leq b \leq \frac{u+\sqrt{u^2-3v}}{3} \leq c \leq \frac{u+2\sqrt{u^2-3v}}{3}\,.$$
We also have $u^2\geq 3v$ and
$$\frac{-2u^3+9uv-2\left(u^2-3v\right)^{3/2}}{27}\leq w \leq \frac{-2u^3+9uv+2\left(u^2-3v\right)^{3/2}}{27}\,.$$
(If $a<b<c$ is assumed, then all inequalities above are strict.)
When $u=6$ and $v=9$, we get
$$0\leq a\leq 1 \leq b\leq 3 \leq c\leq 4\,.$$
Furthermore, $0\leq w \leq 4$ ($w=-d$ in OP's notation).
P.S.  The general case (with $u,v,w$) can also be verified, using the inequalities $(b-c)^2\geq 0$, $(a-b)^2\geq 0$, $(b-a)(c-a)\geq 0$, $(b-a)(c-b)\geq 0$, and $(c-a)(c-b)\geq 0$, as in the first part of this answer.  However, the work is much more tedious than the proof using polynomials as illustrated in the second part of this answer.
A: 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 $c>3$....
