The Floor

$\text{We know that : }\\ \lfloor x \rfloor \leq x \\ \implies \lfloor \dfrac{1}{x^2+1} \rfloor \leq \dfrac{1}{x^2 + 1} \\ \text{The equation } \dfrac{1}{x^2+1} \text{, can never be '0'. That gives us an inequality} \\ 0 < \dfrac{1}{x^2 + 1} \\ \implies \lfloor 0 \rfloor = \lfloor \dfrac{1}{x^2 + 1} \rfloor \implies 0 = \lfloor \dfrac{1}{x^2 + 1} \rfloor$

You can choose any number of $x$ , then subtitute into the equation. I choose $x=1$ , then we get $\left \lfloor \frac{1}{2} \right \rfloor$ . Note that every value of $x$ will leave a fraction with $"1"$ is the numerator and $x^{2}+1 > 1$ . That gives us $0< \frac{1}{x^2+1} <1$ . So, the value of $\left \lfloor \frac{1}{x^{2}+1} \right \rfloor= 0$