Question

√a²+b² + √ b²+c² + √ c²+a² => ( a+b+c)√2 . Prove​

Answer 1

Answer:

If a^2+b^2+c^2-ab-bc-ca=0 then prove that a=b=c.

a² + b² + c² = ab + bc + ca

On multiplying both sides by ‘2’, it becomes

2 ( a² + b² + c² ) = 2 ( ab + bc + ca)

2a² + 2b² + 2c² = 2ab + 2bc + 2ca

a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0

a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0

(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca) = 0

(a – b)² + (b – c)² + (c – a)² = 0

=> Since the sum of square is zero then each term should be zero

⇒ (a –b)² = 0, (b – c)² = 0, (c – a)² = 0

⇒ (a –b) = 0, (b – c) = 0, (c – a) = 0

⇒ a = b, b = c, c = a

∴ a = b = c.