Prove determinant is negative

Question

Prove that the determinant $\Delta$ is negative

$$ \Delta=\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}<0 $$

where $a,b,c$ are positive and $a\neq b\neq c$.

My Attempt:

Applying Sarrus' rule, $$ \begin{matrix} a&b&c&a&b\\ b&c&a&b&c\\ c&a&b&c&a \end{matrix} $$ $$ \Delta=acb+bac+cba-c^3-a^3-b^3=3abc-(a^3+b^3+c^3)\\ =-\Big[(a^3+b^3+c^3)-3abc\Big] $$ How do I prove that $(a^3+b^3+c^3)-3abc>0$ thus prove $\Delta<0$ ?

Answer 1

By the AM-GM inequality:

$$ \frac{a^3+b^3+c^3}{3} \ge \sqrt[3]{a^3b^3c^3} = abc $$

The strict inequality holds unless $\,a=b=c\,$.

Answer 2

You can factor $$a^3 + b^3 + c^3 - 3 abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca),$$ where the first term is certainly positive. For the second term, observe $$a^2 + b^2 + c^2 - ab-bc-ca = \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2),$$ which is also positive.