<s>[INST] 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.
 [/INST] Suppose $r$ is the smallest and $s$ is the largest roots of the denominator. (There are 2 of them because of your second inequality). Set $f(x)=\frac{x+a} {x^2+bx+c^2}$ .Now if $r+a>0$ then $\lim_{r^{+}}f=-\infty$...  Similarly we can prove the case $r+a<0$  and $r+a=0$...
</s>