Question

prove that a square + b square + c square minus A B minus b c minus A is always non negative for all the values of a b c​

Answer 1

HEYA!!

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.

HOPE IT WORKS!!!

Answer 2

here is your answer

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