Show that the roots of the quadratic equation

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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: $b\le c\le a:$ $$(c-a)^2\ge 0 \ge (a-b)(b-c).$$