Functions 02

There are multiple ways to do this, and as I try finding the extreme point, this solution also came up to me. For real number $x$ , $x^2 \geq 0$ , thus: $x^2+1 \geq 1;\: 0 < \frac{1}{x^2+1} \leq 1$ When put $x^2$ in, it will be $x^2< x^2+\frac{1}{x^2+1} \leq x^2+1$ as $1 \leq x^2+1$ , the lowest possible value of $f(x)$ is $1$ from $x=0$