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If a2 + b2 + c2 = 1, then ab + ac + bc lies in the interval
- a)[1, 2/3]
- b)[-1/2, 1]
- c)[–1, 1/2]
- d)[2, –4]
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
If a2+ b2+ c2= 1, then ab + ac + bc lies in the intervala)[1, 2/3]b)[-...
-1, 1/2]
d)[2/3, 1]
e) [-1, -2/3]
We can start by using the AM-GM inequality on the given expression:
ab + ac + bc = a(b+c) + bc = a(-a) + bc = -(a^2 - bc)
Now, we can use the given condition that a^2 + b^2 + c^2 = 1 to write:
a^2 - bc = a^2 + b^2 + c^2 - (b^2 + c^2 - 2bc) = 1 - (b-c)^2
Substituting this back into the original expression, we get:
ab + ac + bc = -(1 - (b-c)^2)
Since (b-c)^2 is always non-negative, we have:
-1 ≤ -(1 - (b-c)^2) ≤ 0
Therefore, the answer is (d) [2/3, 1].
d)[2/3, 1]
e) [-1, -2/3]
We can start by using the AM-GM inequality on the given expression:
ab + ac + bc = a(b+c) + bc = a(-a) + bc = -(a^2 - bc)
Now, we can use the given condition that a^2 + b^2 + c^2 = 1 to write:
a^2 - bc = a^2 + b^2 + c^2 - (b^2 + c^2 - 2bc) = 1 - (b-c)^2
Substituting this back into the original expression, we get:
ab + ac + bc = -(1 - (b-c)^2)
Since (b-c)^2 is always non-negative, we have:
-1 ≤ -(1 - (b-c)^2) ≤ 0
Therefore, the answer is (d) [2/3, 1].
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Community Answer
If a2+ b2+ c2= 1, then ab + ac + bc lies in the intervala)[1, 2/3]b)[-...
Concept:
1. (a - b)2 = a2 - 2ab + b2
2. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Where, (a + b + c)2 > 0
If a, b & c ∀ 0, then
(a + b + c)2 = 0
Calculation:
Given that,
a2 + b2 + c2 = 1 ----(1)
We know that, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ (ab + bc + ca) ≥ -1/2
Again ( b − c )2 + ( c − a )2 + ( a − b )2 ≥ 0
⇒ bc + ca + ab ≤ a2 + b2 + c2 = 1
Hence − 1/2 ≤ bc + ca + ab ≤ 1
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