UPSC NDA Math Model Paper 3

Simplify the given determinant as follows:
|

a	b	c
b	c	a
c	a	b

| =a(bc−a2)−b(b2−ac) +c(ab−c2)
=abc−a3−b3+abc+abc−c3
=3abc−(a3+b3+c3)
As we know, AM ≥ GM
⇒

a3+b3+c3

3

≥3√a3×b3×c3
⇒a3+b3+c3>3abc
Now it is given that a, b and c are distinct positive real numbers. Therefore, the sum of the cube a3+b3+c3>abc, it implies that the value of the determinant is always gonna be negative