Question

Prove that a2b2c2-ab-bc-ca is always non-negative for all valuesof a, b and c.

Answer 1

HEY!!

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✴a2 + b2 + c2 - ab - bc - ca

▶(1/2)(2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca)

▶(1/2)(a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2)

▶(1/2)[(a-b)2 + (b-c)2 + (c-a)2]

▶Now we know that, the square of a number is always greater than or equal to zero.

✔Hence, (1/2)[(a-b)2 + (b-c)2 + (c-a)2] ≥ 0 it is always non-negative.

Answer 2

Prove : -

(a + b + c )^2 = a ^2 +b ^2 + c^2 + 2( ab+ bc + ac)
As the square of any number is always greater than 0

Therefore, 
a^2 +b^2 +c^2 + 2(ab +bc + ca) > 0
a^2 +b^2 +c^2 > - 2 ( ab + bc + ca)
(a^2 +b^2 +c^2)/2 > -ab -bc - ca



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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