If a²+b²+c²=2(b+c-1) then find the value of a+b+c
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Step-by-step explanation:
To solve this question, you must know the identity
(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]
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