Question

A square + b square +c =1 then ab +bc + ac lies in

Answer 1

Answer:

... lies in the interval [ -1/2, 1 ]

Step-by-step explanation:

( a - b )² + ( b - c )² + ( c - a )² ≥ 0

=> a² + b² - 2ab + b² + c² - 2bc + c² + a² ≥ 0

=> 2 ( a² + b² + c² ) ≥ 2 ( ab + bc + ca )

=> 1 ≥ ab + bc + ca

( a + b + c ) ² ≥ 0

=> a² + b² + c² + 2 ( ab + bc + ca ) ≥ 0

=> 2 ( ab + bc + ca ) ≥ -1

=> ab + bc + ca ≥ -1/2

Both of the bounds are attainable.

When ( a, b, c ) = ( 1/√3, 1/√3, 1/√3 ), the upper bound is attained.

When ( a, b, c ) = ( 1/√2, -1/√2, 0), the lower bould is attained.