<s>[INST] If $(y^2-5y+3)\cdot (x^2+x+1)<2x$ for all $x\in \mathbb{R}\;,$ Then Range of $y$ is 
If $(y^2-5y+3)\cdot (x^2+x+1)<2x$ for all $x\in \mathbb{R}\;,$ Then Range of $y$ is

$\bf{My\; Try::}$ We can write the given inequality as $\displaystyle y^2-5y+2<\frac{2x}{x^2+x+1}.$
Now we will calculate range of $$\displaystyle z = \frac{2x}{x^2+x+1}\Rightarrow zx^2+zx+z = 2x$$
So  $$zx^2+x(z-2)+z=0\;,$$ Now if given equation has real roots, Then $\bf{Discriminant \geq 0}$
So  $$(z-2)^2-(2z)^2\geq 0 \Rightarrow (2z)^2-(z-2)^2\leq 0$$  
So $$\displaystyle (z+2)\cdot (3z-2)\leq 0 \Rightarrow (z+2)\cdot \left(z-\frac{2}{3}\right)\leq 0 \Rightarrow -2 \leq z\leq \frac{2}{3}$$
Here  above we have given $$\displaystyle y^2-5y+3<\frac{2x}{x^2+x+1} \leq -2$$
So $\displaystyle y^2-5y+3< -2\Rightarrow y^2-5y+5 < 0$ 
So we get $$\displaystyle  \left(\frac{5+\sqrt{5}}{2}\right)< y < \left(\frac{5+\sqrt{5}}{2}\right)$$
Is my solution is Right, If not then plz explain me , Thanks 
 [/INST] If $x<0$, then by AM-GM, $|x|+\frac{1}{|x|}\geq 2$, whence
$$\frac{2x}{x^2+x+1}=\frac{2}{1-\left(|x|+\frac{1}{|x|}\right)}\geq \frac{2}{1-2}=-2\,.$$
The equality holds iff $x=-1$.  If $x\geq 0$, then $\frac{2x}{x^2+x+1}\geq 0>-2$.  That is, $y^2-5y+3<-2$, meaning that all possible values of $y$ are $y\in\mathbb{R}$ such that $$\frac{5-\sqrt{5}}{2}<y<\frac{5+\sqrt{5}}{2}\,.$$
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