<s>[INST] 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?
 [/INST] 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.
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