Question

If a square + b square + c square is equal to 14 then a b + BC + CA is always greater than or equal to

Answer 1

We know (a+b+c)^2= a^2+b^2+c^2+2(ab+bc+ca)

Now, putting the given value

(a+b+c)^2= 1+2(ab+bc+ca)

2(ab+bc+ca)> -1 since (a+b+c)^2 cannot be less than zero

or ab+bc+ca > -1/2 for all real value of a, b, c.

Answer 2

Thank you for asking this question. Here is your answer:

(a+b+c)^2

= a^2 + b^2 + c^2 + 2(ab + bc + ca)

(a+b+c)^2 ≥ 0

a^2 + b^2 + c^2 + 2(ab + bc + ca) ≥ 0

a^2 + b^2 + c^2 = 1 (this is given)

1 + 2(ab + bc + ca) ≥ 0

ab + bc + ca ≥ -1/2

(a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0

2 [ a^2 + b^2 + c^2 - ab - bc - ca ] ≥ 0

2 [ 1 - (ab + bc + ca)] ≥ 0

1 ≥ (ab + bc + ca)

So the final answer for this question is: [-1/2, 1]

If there is any confusion please leave a comment below.