<s>[INST] If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be? If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be?
$$y^2(x^2-5x+4)+y(x^2-5x+2)+(x^2-5x+4)<0$$
As it is true for all real y, hence $D<0$
$$(x^2-5x+2)^2-4(x^2-5x+4)^2<0$$
Let $x^2-5x=u$
$$(u+2)^2-4(u+4)^2<0$$
$$u^2+4+4u-4(u^2+16+8u)<0$$
$$-3u^2-28u-60<0$$
$$3u^2+28u+60>0$$
$$3u^2+18u+10u+60>0$$
$$(3u+10)(u+6)>0$$
Back substituting u
$$(3x^2-15x+10)(x^2-5x+6)>0$$
$$\left(x-\dfrac{15+\sqrt{225-120}}{6}\right)\left(x-\dfrac{15-\sqrt{225-120}}{6}\right)(x-2)(x-3)>0$$
$$x\in\left(-\infty,\dfrac{15+\sqrt{105}}{6}\right)\cup(2,3)\cup\left(\dfrac{15+\sqrt{105}}{6},\infty\right)$$
But in the question it is given $x\in(2,b)$, what mistake am I making here.
 [/INST] From $$y^2+y+1=(y+\tfrac12)^2+\tfrac34>0,$$
we know that we may divide by $y^2+y+1$. 
What can we say about $\frac{2y}{y^2+y+1} $?
For $y\ge0$, this will be positive, but what about negative $y$?
We have  $$y^2+y+1=(y-1)^2-y\ge-y$$
with equality iff $y=-1$. Hence $$\frac{2y}{y^2+y+1}\ge-2 $$with equality iff $y=-1$.
So what you really want is
$$x^2-5x+4<-2. $$
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