Plz solve the problem
Answers
Hey there !
Solution:
We know that, for any square of a number, the value is always greater than 0.
=> ( a + b + c )² ≥ 0
=> ( a² + b² + c² + 2ab + 2bc + 2ca ) ≥ 0
Given that, ( a² + b² + c² ) = 1, Hence we get,
=> 1 + 2 ( ab + bc + ac ) ≥ 0
=> 2 ( ab + bc + ac ) ≥ -1
=> ( ab + bc + ac ) ≥ - 1 / 2 => Equation 1
Now get the minimum value we can calculate,
( a - b )² + ( b - c )² + ( a - c )² ≥ ?
=> ( a² + b² - 2ab ) + ( b² + c² - 2bc ) + ( a² + c² - 2ac ) ≥ 0
=. 2a² + 2b² + 2c² - 2ab - 2bc - 2ac ≥ 0
=> 2 ( a² + b² + c² - ab - bc -ac ) ≥ 0
=> 2 [ 1 - ( ab + bc + ca ) ] ≥ 0
=> 1 - ( ab + bc + ca ) ≥ 0 [ ∵ 0 / 2 = 0 ]
=> - ( ab + bc + ac ) ≥ -1
Multiplying by -1 on both sides we get,
=> ab + bc + ac ≤ 1
[ Sign gets inverted as we are multiplying a term on both sides ]
Hence we get,
=> ab + bc + ac = [ - 1/2 , 1 ]
Hence option ( a ) is correct.
Hope my answer helped !