If $a^2+c^2>ab$ and $b^2>4c^2$ , for real x, show that $\frac{x+a}{x^2+bx+c^2}$ cannot lie between two limits If $a^2+c^2>ab$ and $b^2>4c^2$ , for real x, show that $\frac{x+a}{x^2+bx+c^2}$ cannot lie between two limits
My attempt is as follows:
$$y=\frac{x+a}{x^2+bx+c^2}$$
$$yx^2+byx+yc^2=x+a$$
$$yx^2+x(by-1)+yc^2-a=0$$
As x is real,so
$$D>=0$$
$$(by-1)^2-4y(yc^2-a)>=0$$
$$b^2y^2+1-2by-4y^2c^2+4ay>=0$$
$$(b^2-4c^2)y^2+2(2a-b)y+1>=0$$
As it is given $b^2>4c^2$, it means parabola is upwards, now this parabola will not lie between two limits if it does not cut x-axis at two distinct points.
So if $D<=0$ then parabola $(b^2-4c^2)y^2+2(2a-b)y+1>=0$ will not cut x-axis at two distinct points.
So lets calculate D for the equation $(b^2-4c^2)y^2+2(2a-b)y+1=0$
$$D=4(4a^2+b^2-4ab)-4(b^2-4c^2)$$
$$D=4(4a^2+4c^2-4ab)$$
$$D=16(a^2+c^2-ab)$$
But I am getting $D>0$ as $a^2+c^2>ab$
I am getting totally reversed result. What mistake am I doing here. Please help me.

$x,y$ are reak.
Let $$y=\frac{x+a}{x^2+bx+c^2}~~~~(1)$$. Due to reality of $x$ we impose $B^2 \ge 4AC$ on the quadratic of $x$ from (1). We get
$$(b^2-4c^2) y^2 +y(4a-2b)+1 \ge 0, \forall ~ y\in R.~~~(2) $$
A quadratic $Az^2+Bz+C \ge 0, \forall Z \in R,~ then~. A>0$ and $B^2\le 4AC$ Applying this
to the quadratic (2). Hence, we get $$b^2-4c^2 >0~~~(3)$$  and  $$4(2a-b)^2 \ge 4(b^2-4c^2) \implies (a^2+c^2) \ge ab. ~~~(4)$$ This means $y$ can take any real positive or negative.
This is also clear from the numerator of $y$ in (1) which is bound to have two real roots as $b^2 >4c^2$ at near thse roots $y$ will take any real value implying that $y$ is un-bounded not essentially lying between two values.
