atbrc = 14 -Find a b c

1.	atbrc = 14 -Find a b c
Answer» (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 = 1Therefore,1 + 2(ab + bc + ca) ≥ 0 ab + bc + ca ≥ -1/2 (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 02 [ a^2 + b^2 + c^2 - ab - bc - ca ] ≥ 02 [ 1 - (ab + bc + ca)] ≥ 0Therefore, 1 ≥ (ab + bc + ca) Hence the answer is [-1/2, 1]

Answer»

(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 = 1Therefore,1 + 2(ab + bc + ca) ≥ 0

ab + bc + ca ≥ -1/2

(a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 02 [ a^2 + b^2 + c^2 - ab - bc - ca ] ≥ 02 [ 1 - (ab + bc + ca)] ≥ 0Therefore, 1 ≥ (ab + bc + ca)

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

Discussion