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.

Notice that $by=\frac{ax+cz}{2}$, so that
$$
acy^2 \geq b^2y^2=\bigg(\frac{ax+cz}{2}\bigg)^2 \geq ax\times cz 
$$
and hence $y^2 \geq  xz$ (notice that $ac>0$ because $ac>b^2$).
