Solution

Applying \(C_1 \to C_1 + C_2 + C_3\), we have \(\Delta = |\begin{smallmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{smallmatrix}| = (a+b+c)|\begin{smallmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{smallmatrix}|\) Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\), we get \(\Delta = (a+b+c)|\begin{smallmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{smallmatrix}|\) \(= (a+b+c)[(c-b)(b-c) - (a-c)(a-b)] = (a+b+c)[-c^2+2bc-b^2 - (a^2-ab-ac+bc)]\) \(= (a+b+c)(-c^2-b^2-a^2+ab+bc+ca) = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\) \(= -\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]\) Given that a, b, c are positive and unequal. Therefore \(a+b+c > 0\) and \((a-b)^2 + (b-c)^2 + (c-a)^2 > 0\). Hence \(\Delta < 0\).