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.
 A: 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 $ac>b^2$).
A: It seems user2838619 and Ewan Delanoy 
already gave correct solutions with elementary inequalities.
Still the following viewpoint may be of interest, connecting the result with the
Schur
product theorem.
Assume on the contrary that $zx-y^2 > 0$.
The inequality $ac-b^2$ is equivalent to the statement
that the symmetric matrix $M := \left({a \; b \atop b \; c}\right)$ has
positive determinant, and thus that either $M$ or $-M$ is positive definite.
Likewise $zx-y^2 > 0$ means either $N := \left({x \; y \atop y \; z}\right)$
or $-N$ is positive definite.  By Schur it follows that Hadamard product
$$
M \circ N = \left( {ax \ \ by \atop by \ \ cz} \right)
$$
is either positive or negative definite.  This contradicts the hypothesis
$ax-2by+cz = 0$, because
$$
ax - 2by + cz = (1,-1) \left( {ax \ \ by \atop by \ \ cz} \right) (1,-1)^T
= \big\langle (1,-1), \ (M\circ N) (1,-1)^T \big\rangle.
$$
A: 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 $\displaystyle\implies b^2y^2\ge aczx>b^2zx\iff b^2(y^2-zx)>0\iff y^2>zx$
A: Consider the following equation.
$$axt^2-2byt+cz=0.$$
Since $ax-2by+cz=0$, we see that this equation has root $1$, which gives
$$b^2y^2-acxz\geq0.$$
Since $ac>b^2$, we obtain $ac>0$ and since
$$acy^2\geq b^2y^2\geq aczx,$$ we obtain
$$acy^2\geq aczx,$$
which gives $y^2\geq zx$.
Done!
