Range over some other range!

Since the expression $y=\frac{2x}{1+x^{2}}$ can take any real value of $x$ , we assume:

$x=\tan \theta \qquad \left[ \text{ where}\ \theta\in \mathbb{R}-(2n+1)\frac { \pi }{ 2 } \right]$

Putting the above value of $x$ in the expression of $y$ we get:

$y=\frac{2 \tan \theta}{1+\tan \theta^{2}} \Rightarrow y=\sin 2\theta$

$\Rightarrow y \in [-1,1]$

Let the given expression be $E$ . And now playing further with it:

$\begin{array}{ll} E & =y^2+y-2 \\ & =\left( \sin 2 \theta +\frac{1}{2} \right)^2 -\frac{9}{4}\end{array}$

Note : ${(Anything)^2 >= 0}$

$E$ will gain the minimum value when $\sin 2\theta=-\frac 12$ at which $E$ will be $-\frac 94$ . And it will gain the maximum value when $\sin 2\theta=1$ at which $E$ will be $0$ .

Hence the range of $E$ is $\left[-\frac { 9 }{ 4 } ,0 \right].$

NOTE: One can also use AM-GM Inequality by dividing numerator and denominator by $x(\neq 0)$ and taking two case : $x > 0$ and $x<0$ and then proceed with AM-GM inequality.