Question

• If 3x - 7y
prove that a² + b^2 + c^2 - ab - bc - ca is always non-negative for all values of a, b and c​

Answer 1

Step-by-step explanation:

We have,

a² + b² + c² − ab − bc − ca

Multiply and divide by ‘2’

= 2/2[a² + b² + c² − ab − bc − ca]

= 1/2[2a² + 2b² + 2c² − 2ab − 2bc − 2ca]

= 1/2[a² + a² + b² + b² + c² + c² − 2ab − 2bc − 2ca]

= 1/2[(a² + b² − 2ab) + (a² + c² − 2ca) + (b² + c² − 2bc)]

= 1/2[(a − b)² + (b − c)² + (c − a)²] [(a − b)² = a² + b² − 2ab]

∴ a² + b² + c² − ab − bc − ca ≥ 0

Hence, a² + b² + c² − ab − bc − ca ≥ 0 is always non-negative for all values of a, b and c.

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