Question

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Prove that a^2+b^2+c^2-ab-bc-ca is always negative for all values of ab and c.

Answer 1

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Here's , ur answer :-

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A2 + 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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