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How to proof that a^2+b^2+c2-ab-bc-ca is negative for all values of a,b,and c

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2 years ago

```							FIrst of all this value be non negative not negativea2 + b2 + c2 − ab − bc − ca Multiplying and dividing the expression by 2, = 2(a2 + b2 + c2 − ab − bc – ca) / 2 = (2a2 + 2b2 + 2c2 − 2ab − 2bc − 2ca) / 2 = (a2 − 2ab + b2 + b2 − 2bc + c2 + c2 − 2ca + a2) / 2 = [(a − b)2 + (b − c)2 + (c −  a)2] / 2 Square of a number is always greater than or equal to zero. Hence sum of the squares is also 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}/2= 0 when a = b = c. Hence, a2 + b2 + c2 – ab – ac - bc is always non-negative for all its values of a, b and c.
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2 years ago
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