Question

If (a square+ b square+ c square)= (b+c-1) then find a+b+c​

Answer 1

Step-by-step explanation:

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

(a+b+c)^2 ≥ 0 for any real values of a, b, c

Therefore,

a^2 + b^2 + c^2 + 2(ab + bc + ca) ≥ 0

Given that a^2 + b^2 + c^2 = 1

Therefore,

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

Therefore, 1 ≥ (ab + bc + ca)

Hence the answer is [-1/2, 1]