Calculus approach to solve this Quadratic equation problem 
Both roots of the equation $$(x-b) (x-c) +(x-a) (x-c) +(x-a) (x-b) = 0$$ are always positive , negative or real. Prove your result. 

By solving this equation I got $3x^2 - 2(a+b+c)x +ab + bc + ca = 0$
Now by inspection of this quadratic : 
$a>0$ and $D = 4(a^2 + b^2 + c^2 -ab - bc - ca)$
I'm stuck here and unable to conclude whether $D>0$ , $D<0$ or $D = 0$
Also is there an alternative approach to solve these types of questions. Kindly suggest any other method (if any) too.
By various answers here now I know where I was stuck. But I just want to know alternate method to approach this question without having to do this much algebra.
 A: Edit : I divide this answer into 2 parts. The first one originates as a quick answer to the question; the second one has my preference because it provides an intuitive understanding of the situation.
Part 1 : 
With your notations ($D$ being the discriminant):
$$\dfrac{D}{4} = a^2 + b^2 + c^2 -ab - bc - ca=\dfrac{1}{4}(2a-b-c)^2+\dfrac{3}{4}(b-c)^2$$
Thus, in general $D\geq 0$ with:


*

*$D=0$ when simultaneously $b=c$ and $2a-b-c=0$, which means if and only if all $a=b=c$. (case of a single root).

*$D>0$ otherwise. (two real roots).
Part 2 : 
Before beginning we assume that $a,b,c$ are distinct and WLOG that $a<b<c$.
Dividing by $(x-a)(x-b)(x-c)$, we have the following equivalent equation .
$$\dfrac{1}{x-a}+\dfrac{1}{x-b}+\dfrac{1}{x-c}=0  \ \ \ (1)$$
In fact, set $f(x)$ for the LHS of (1); it is immediate that f'(x)<0, thus $f$ is a strictly decreasing function (see graphical representation example below). Consider interval $(a,b)$: $lim_{x \rightarrow a+0} f(x) =+\infty$ and $lim_{x \rightarrow b-0} f(x) =-\infty$. Thus there is a root $x_1$ in this interval. Besides, this root is unique because $f$ is strictly decreasing. The same for interval $(b,c)$.
Thus we have two real roots such that:
$$a<x_1<b<x_2<c$$
No more than two real roots may occur as we know from the first form of the equation.

A: $$f(x)=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$$
Then,
$$f(a)=(a-b)(a-c)$$
$$f(b)=(b-c)(b-a)$$
$$f(c)=(c-a)(c-b)$$
Clearly, the coefficient of $x^2$ in $f(x)$ is positive. Suppose there are no real roots. Then, $f(x)\gt0 \forall x\in\mathbb R$.
$$(a-b)(a-c)\gt0$$
Without loss of generality, let $a\gt b$ and $a\gt c$.
$$(b-c)(b-a)\gt 0\Rightarrow c\gt b$$
$$(c-a)(c-b)\gt 0\Rightarrow b\gt c$$
This leads to a contradiction. Thus, the graph must meet the $x$-axis and thus the roots are real.
