If $a>b>c$ and the system of equations $ax+by+cz = 0, bx + cy + az = 0$ and $cx+ay + bz=0$ has a non-trivial solution, then the quadratic equation $ax^2 + bx + c = 0$ has

at least one positive root

The given system of equations has a non-trivial solution

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

$⇒(a+b+c) (a^2 + b^2 + c^2-ab-bc-ca) = 0$

$⇒(a+b+c) [(a-b)^2 + (b −c)^2 + (c-a)^2]=0$

$⇒a+b+c=0$    $[∵ (a-b)^2 + (b −c)^2 + (c-a)^2 ≠0]$

$⇒x = 1$ is a root of $ax^2 + bx + c = 0$

Thus, the quadratic equation has at least one positive root.