Show that the roots of the quadratic equation 
Show that the roots of the quadratic equation $$(b-c)x^2+2(c-a)x+(a-b)=0$$
  are always reals.

My Attempt:
$$(b-c)x^2+2(c-a)x+(a-b)=0$$
Comparing above equation with $Ax^2+Bx+C=0$
$$A=b-c$$
$$B=2(c-a)$$
$$C=(a-b)$$
Now,
$$B^2-4AC=[2(c-a)]^2-4(b-c)(a-b)$$
$$=4(c-a)^2 - 4(ab-b^2-ac+bc)$$
$$=4c^2-8ac+4a^2-4ab+4b^2+4ac-4bc$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
How do I proceed further?
 A: You need to prove that
$$a^2+b^2+c^2\ge ab+bc+ca.$$
This is a well-known inequality, and can be deduced from the
two-variable AM/GM inequality, which states
$$a^2+b^2\ge2ab.$$
A: The first part of your solution is true.
We can end it by the following way.
$b-c\neq0$ by given and we need to prove that
$$(c-a)^2-(b-c)(a-b)\geq0$$ or
$$a^2+b^2+c^2-ab-ac-bc\geq0$$ or
$$\sum_{cyc}(a^2-ab)\geq0$$ or
$$\sum_{cyc}(2a^2-2ab)\geq0$$ or
$$\sum_{cyc}(a^2-2ab+b^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2\geq0,$$
which is obvious.
A: Certainly, using the AM-GM is easier. An alternative method is to consider cases. We want to prove the condition $D\ge 0$ for all $a,b,c\in R$:
$$D=(c-a)^2-(c-b)(a-c)\ge 0 \Rightarrow \\ (c-a)(c-a)\ge(a-b)(b-c)=(b-a)(c-b).$$
Case 1: $a\le b\le c:$
$$c-a\ge b-a; \ c-a\ge c-b.$$
Case 2: $a\le c\le b:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
Case 3: $c\le a\le b:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
Case 4: $c\le b\le a:$
$$(c-a)^2=(a-c)^2 \ge (a-b)(b-c).$$
Case 5: $b\le a\le c:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
Case 6: $b\le c\le a:$
$$(c-a)^2\ge 0 \ge (a-b)(b-c).$$
