Question

if a²+b²+c²=2(b+c-1),then find the value of a+b+c​

Answer 1

Answer:

To solve this question, you must know the identity

(a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

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

Therefore

a² + b² + c² + 2(ab + bc + ca) ≥ 0

Given that a² + b² + c² = 1

Therefore

1 + 2(ab + bc + ca) ≥ 0

ab + bc + ca ≥ -1/2

(a-b)² + (b-c)² + (c-a)² ≥ 0

2 [ a² + b² + c² - 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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Step-by-step explanation:

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