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Prove that A2 + B2 + C2 − Ab − Bc − Ca is Always Non-negative for All Values of A, B and C - Mathematics

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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

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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Chapter 4: Algebraic Identities - Exercise 4.1 [Page 7]

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RD Sharma Mathematics [English] Class 9
Chapter 4 Algebraic Identities
Exercise 4.1 | Q 14 | Page 7

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