Suppose a, b, c are sides of a scalene triangle. Let Δ=a b cb c ac a b .Then
Δ≤0
Δ<0
Δ>0
Δ≥0
Using C1→C1+C2+C3 we get
Δ=(a+b+c)Δ1
where
Δ1=1 b c1 c a1 a b
Applying R2→R2−R1,R3→R3−R1 we get
Δ1=1bc0c−ba−c0a−bb−c=−(b−c)2−(a−b)(a−c)=−a2+b2+c2−bc−ca−ab
⇒ Δ1=−12(b−c)2+(c−a)2+(a−b)2<0
As a+b+c>0, we get
Δ=(a+b+c)Δ1<0