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Question
Prove that a2 + b2 + c2 − ab − bc − ca is always non-negative for all values of a, b and c
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Solution
We have
`a^2 + b^2 + c^2 - ab - bc - ca`
`= 2/2[a^2 + b^2 + c^2 - ab - bc - ca]` [Mulitply and divide by 2]
`= 1/2 [2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca]`
`= 1/2 [a^2 + a^2 + b^2 + b^2 + c^2 - 2ab - 2bc - 2ac]`
`= 1/2[(a^2 + b^2 - 2ab) + (a^2 + c^2 - 2ac) + (b^2 + c^2 - 2bc)]`
`= 1/2 [(a - b)^2 + (b - c)^2 + (c - a)^2]` `[∵ (a - b)^2 = a^2 + b^2 - 2ab]`
`= ((a - b)^2 + (b -c)^2 + (c - a)^2)/2 >= 0`
`∴ a^2 + b^2 + c^2 - ab - bc -ca >= 0`
hence `a^2 + b^2 - ab - bc - ca > 0`
Hence `a^2 + b^2 + c^2 - ab - bc - ca` is always non-negative for all values of a, b and c.
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