Brove that a + b ² + c² - ab-cb-ca is
always con segalue for all value of a, b
and C
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Answer:
2
+b
2
+c
2
−ab−bc−ca
multiply and divide by 2
=
2
2
×(a
2
+b
2
+c
2
−ab−bc−ca)
=
2
a
2
−2ab+b
2
+b
2
−2bc+c
2
+c
2
−2ac+a
2
=
2
(a−b)
2
+(b−c)
2
+(c−a)
2
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 when a=b=c
Hence, a
2
+b
2
+c
2
−ab−bc−ca is always non negative
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