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?

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
$$a^2+b^2\ge2ab.$$
