(a-b)

^{2}+(b-c)^{2}+(c-a)^{2}>=0

How did you come up with this equation?

10 votes

18 votes

Best answer

**Ans (B) **

$(a+b+c)^2$ should be $\geq 0.$

$ \therefore a^2 + b^2 +c^2 + 2ab +2bc + 2ca \geq 0$

Given $a^2 + b^2 +c^2 = 1$ therefore $1+ 2(ab+bc+ca)\geq 0$

$\implies ab+bc+ca\geq -\frac{1}{2}$

Now to find the upper limit,

$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$

Expanding we get, $2(a^2 + b^2 + c^2) - 2(ab+bc+ca) \geq 0$

$\implies 2-2(ab+bc+ca)\geq 0$

$\implies ab+bc+ca\leq 1$

**Answer B. [-1/2,1] **

1

@Tuhin Dutta It is the sum of squares. Therefore, it will never be negative. It will either be positive or zero.