Question

Prove that a^2 + b^2 + c^2 - ab - ac - bc is always non-negative for all the values of a,b and c

Answer 1

Hi there!

Here's your answer :-

a² + b² + c² − ab − bc − ca
 
Multiply n' divide th' 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
 
We kn'w
Square of a number is always greater than or equal to zero.
 
Therefore, sum of the squares is also greater than or equal to zero
 
∴ [ (a − b)² + (b − c)² + (c − a)² ] ≥ 0  

n'  { (a − b)² + (b − c)² + (c − a)² } / 2 = 0 when a = b = c.
 
Hence, The required answer is :-

a² + b² + c² – ab – ac - bc is always non-negative for all its values of a, b and c.

Hope it helps!

Answer 2

Answer:

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