if a,b and c are distinct positive integers then the least value of ( a²+b²+c²-ab-bc-ca) is equal to
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Answer:
a
2
+b
2
+c
2
−(bc+ca+ab)
=
2
1
(2a
2
+2b
2
+2c
2
−2bc−2ca−2ab)
=
2
1
((b
2
+c
2
−2bc)+(c
2
+a
2
−2ca)+(a
2
+b
2
−2ab))
=
2
1
((b−c)
2
+(c−a)
2
+(a−b)
2
)>0
⇒1−(bc+ca+ab)>0⇒bc+ac+ab>1
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Answered by
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Given,
a, b and c are distinct positive integers.
To find,
The value of ( a²+b²+c²+ab-bc-ca)
Solution,
According to the question,
( a²+b²+c²+ab-bc-ca)
Multiply & divide the equation by 2
=2/2× (a²+b²+c²+ab-bc-ca)
=( a²-2ab+b²+b²-2bc+c²+c²-2ac+a²)/2
=(a-b)² + (b-c)² + (c-a)²/2
The square of a number is always greater than or equal to zero.
∴ (a-b)² + (b-c)² + (c-a)² ≥ 0
and
(a-b)² + (b-c)² + (c-a)² = 0
Hence, ( a²+b²+c²-ab-bc-ca) = 0.
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