827 views

if $a^2+b^2+c^2=1$ then $ab+bc+ac$ lies in the interval

1. $[1,2/3]$
2. $[-1/2,1]$
3. $[-1,1/2]$
4. $[2,-4]$

edited | 827 views

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, $2(a^2 + b^2 + c^2) - 2(ab+bc+ca) \geq 0$
$\implies 2-2(ab+bc+ca)\geq 0$
$\implies ab+bc+ca\leq 1$

by (453 points)
edited by
0

(a-b)2+(b-c)2+(c-a)2 >=0

How did you come up with this equation?

+1

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