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

$(0,1]$

We have,

$f(x)=\left(1+b^2\right) x^2+2 b x+1$

$\Rightarrow f(x)=\left(1+b^2\right)\left\{x^2+\frac{2 b}{1+b^2} x+\frac{1}{1+b^2}\right\}$

$\Rightarrow f(x)=\left(1+b^2\right)\left\{\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{\left(1+b^2\right)^2}+\frac{1}{\left(1+b^2\right)}\right\}$

$f(x)=\left(1+b^2\right)\left\{x+\frac{b}{1+b^2}\right\}^2+\frac{1}{\left(1+b^2\right)}$

It is evident from this that the minimum value of f(x) is $\frac{1}{1+b^2}$ which it attains at $x=-\frac{b}{1+b^2}$

∴  $m(b)=\frac{1}{1+b^2}$

Clearly, $0<m(b) \leq 1$

Hence, range of $m(b)$ is $(0,1]$