Let  $f\left(x\right)=(1+b^2)x^2+2bx+1$  and let   $m\left(b\right)$  be the minimum value of $f\left(x\right)$.  As b varies, the range of
$m\left(b\right)$  is

$$\begin{gathered} f\left(x\right)=\left(1+b^2\right)x^2+2bx+1 \\ \text{ It is quadratic expression with coefficient of }{x}^2\text{ as }\left(1+b^2\right)>0 \\ \text{ Whose minimum value is }-\frac{\mathrm{D}}{4\mathrm{a}} \\ \therefore \mathrm{m}\left(\mathrm{b}\right)=-\frac{\left\{4{\mathrm{b}}^2-4\left(1+{\mathrm{b}}^2\right)\right\}}{4\cdot \left(1+{\mathrm{b}}^2\right)} \\ m\left(b\right)=\frac{1}{1+b^2}\text{, Thus }m\left(b\right)=(0,1] \\ \mathrm{y}=\frac{1}{1+{\mathrm{b}}^2} \\ \text{ Hence, }0<\frac{1}{1+b^2}\le 1 \\ \Rightarrow \text{ range of }\mathrm{m}\left(\mathrm{b}\right)\text{ is }(0,1] \end{gathered}$$