Step 1:Let $\Delta=\begin{vmatrix}a & b& c\\b & c & a\\c & a & b\end{vmatrix}$
Apply $C_1\rightarrow C_1+C_2+C_3$
Hence $\Delta=\begin{vmatrix}a+b+c & b & c\\a+b+c & c & a\\a+b+c & a & b\end{vmatrix}$
Take (a+b+c) as the common factor from $C_1$
Therefore $\Delta=\begin{vmatrix}1 & b& c\\1 & c & a\\1 & a & b\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$
$\Delta=(a+b+c)\begin{vmatrix}1 & b & c\\0 & c-b & a-c\\0 & a-b & b-c\end{vmatrix}$
Step 2:Now let us expand along $C_1$
$\Delta=(a+b+c)[(c-b)(b-c)-(a-c)(a-b)]$
$\quad=(a+b+c)(-a^2-b^2-c^2+ab+bc+ca)$
Now multiply and divide by -2
$\Delta=-\frac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)$
But we know$(a+b)^2=a^2+2ab+b^2$
$\Delta=-\frac{1}{2}(a+b+c)(a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac)$
$\quad=-\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$
This is negative,since a+b+c>0 and $(a-b)^2+(b-c)^2+(c-a)^2>0.$
Hence proved.