GATE2015 EC-2: GA-9

Question

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

• A. $[1,2/3]$
• B. $[-1/2,1]$
• C. $[-1,1/2]$
• D. $[2,-4]$ 

Answer 1

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]

Comments

(a-b)²+(b-c)²+(c-a)² >=0 

How did you come up with this equation? 

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

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For maximum limit:

You can use the fact that AM >= GM and this will also the above question