if $ax-2by+cz=0$ and $ac-b^2>0$ , Prove $zx-y^2\leq0$ 
for real numbers like $a,b,c,x,y,z$ that $ax-2by+cz=0$ and $ac-b^2>0$ Prove:$$zx-y^2\leq0$$
Additional info: The Proof should be by contradiction.we can use Cauchy , AM-GM and other simple inequalities.

Things I have done so far: as Problem wants a Proof by contradiction, I assume that $zx-y^2>0$ and later show that it is impossible.We know that $ac-b^2>0$. So $$ac+zx-(b^2-y^2)>0$$
by AM-GM we know that $$b^2+y^2\geq 2by$$
So $$ac+zx-2by>0$$
and by $ax-2by+cz=0$ we can write $$ac+zx-ax-cz>0$$
and I stuck here. and another problem about my uncompleted proof is using AM-GM.because the question did not mentioned about being positive real number.it just said real numbers.

If $x,y$ are of opposite sign, $y^2>zx,$ and we are done
Again $ac-b^2>0\iff ac>b^2\ge0\implies a,c$ are of same sign
So, we need to consider $a,c$  & $x,z$ in pair are same sign
Case $(1):$ $a,c>0; z,x>0$
Case $(4):$ $a,c<0; z,x<0$
$$2by=ax+cz\ge2\sqrt{aczx}\implies b^2y^2\ge aczx>b^2zx\iff b^2(y^2-zx)>0\iff y^2>zx$$
Case $(2):$ $a,c<0; z,x>0$
Case $(3):$ $a,c>0; z,x>0$
$$2by=ax+cz\le2\sqrt{aczx}$$
As $ax,cz<0,$
this $\displaystyle\implies b^2y^2\ge aczx>b^2zx\iff b^2(y^2-zx)>0\iff y^2>zx$
