Question

If e, f and g are distinct positive +ve real numbers and e² + f² + g² = 1 then ef + fg + ge is 

1. = 1
2. > 1
3. < 1
4. can not be Determined

Answer 1

Correct Answer - Option 3 : < 1

Calculation:

⇒ (e - f)² + (f – g)² + (g - e)² > 0

⇒ 2 (e² + f² + g² – ef – fg – ge) > 0

⇒ e² + f² + g² – ef – fg – ge > 0

⇒ e² + f² + g²  >  ef + fg + ge

⇒ ef + fg + ge < 1